第2章 经典力学回顾
2.1 最小作用量原理与Lagrange力学
习题 2.1.1 考虑下面这个称为谐振子 的系统。物块的质量为 m m m k k k
解答:
动能和势能分别为
T = 1 2 m x ˙ 2 V = 1 2 k x 2 \begin{aligned}
T&=\frac{1}{2}m\dot{x}^{2}\\
V&=\frac{1}{2}kx^{2}
\end{aligned} T V = 2 1 m x ˙ 2 = 2 1 k x 2 于是Lagrange量为
L = T − V = 1 2 m x ˙ 2 − 1 2 k x 2 \mathscr{L}=T-V=\frac{1}{2}m\dot{x}^{2}-\frac{1}{2}kx^{2} L = T − V = 2 1 m x ˙ 2 − 2 1 k x 2 我们可以计算
∂ L ∂ x ˙ = m x ˙ ∂ L ∂ x = − k x \begin{aligned}
\frac{\partial\mathscr{L}}{\partial \dot{x}}&=m\dot{x}\\
\frac{\partial \mathscr{L}}{\partial x}&=-kx
\end{aligned} ∂ x ˙ ∂ L ∂ x ∂ L = m x ˙ = − k x 因此，Euler-Lagrange方程为
d d t ( ∂ L ∂ x ˙ ) − ∂ L ∂ x = 0 \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \mathscr{L}}{\partial\dot{x}}\right)-\frac{\partial\mathscr{L}}{\partial x}=0 d t d ( ∂ x ˙ ∂ L ) − ∂ x ∂ L = 0 运动方程为
m x ¨ + k x = 0 m\ddot{x}+kx=0 m x ¨ + k x = 0 ■ ~\tag*{$\blacksquare$} ■ 习题 2.1.2
对第 1.8 节末尾讨论的耦合质量问题做同样的计算。将所得运动方程与式 (1.8.24) 和 (1.8.25) 比较。
解答:
系统的动能和势能分别为
T = 1 2 m x ˙ 1 2 + 1 2 m x ˙ 2 2 V = 1 2 k x 2 + 1 2 k ( x 2 − x 1 ) 2 + 1 2 k x 2 2 \begin{aligned}
T&=\frac{1}{2}m\dot{x}_{1}^{2}+\frac{1}{2}m\dot{x}_{2}^{2}\\
V&=\frac{1}{2}kx^{2}+\frac{1}{2}k(x_{2}-x_{1})^{2}+\frac{1}{2}kx_{2}^{2}
\end{aligned} T V = 2 1 m x ˙ 1 2 + 2 1 m x ˙ 2 2 = 2 1 k x 2 + 2 1 k ( x 2 − x 1 ) 2 + 2 1 k x 2 2 于是Lagrange量为
L = T − V = 1 2 m ( x ˙ 1 2 + x ˙ 2 2 ) − k ( x 1 2 − x 1 x 2 + x 2 2 ) \mathscr{L}=T-V=\frac{1}{2}m(\dot{x}_{1}^{2}+\dot{x}_{2}^{2})-k(x_{1}^{2}-x_{1}x_{2}+x_{2}^{2}) L = T − V = 2 1 m ( x ˙ 1 2 + x ˙ 2 2 ) − k ( x 1 2 − x 1 x 2 + x 2 2 )
对 1 1 1
∂ L ∂ x ˙ 1 = m x ˙ 1 ∂ L ∂ x 1 = − 2 k x 1 + k x 2 \begin{aligned}
\frac{\partial\mathscr{L}}{\partial \dot{x}_{1}}&=m\dot{x}_{1}\\
\frac{\partial \mathscr{L}}{\partial x_{1}}&=-2kx_{1}+kx_{2}
\end{aligned} ∂ x ˙ 1 ∂ L ∂ x 1 ∂ L = m x ˙ 1 = − 2 k x 1 + k x 2 d d t ( ∂ L ∂ x ˙ 1 ) − ∂ L ∂ x 1 = 0 \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \mathscr{L}}{\partial\dot{x}_{1}}\right)-\frac{\partial\mathscr{L}}{\partial x_{1}}=0 d t d ( ∂ x ˙ 1 ∂ L ) − ∂ x 1 ∂ L = 0 得到运动方程
m x ¨ 1 + 2 k x 1 − k x 2 = 0 m\ddot{x}_{1}+2kx_{1}-kx_{2}=0 m x ¨ 1 + 2 k x 1 − k x 2 = 0
x ¨ 1 = − 2 k m x 1 + k m x 2 (2.1) \ddot{x}_{1}=-\frac{2k}{m}x_{1}+\frac{k}{m}x_{2}\tag{2.1} x ¨ 1 = − m 2 k x 1 + m k x 2 ( 2.1 )
对 2 2 2
∂ L ∂ x ˙ 2 = m x ˙ 2 ∂ L ∂ x 1 = k x 1 − 2 k x 2 \begin{aligned}
\frac{\partial\mathscr{L}}{\partial \dot{x}_{2}}&=m\dot{x}_{2}\\
\frac{\partial \mathscr{L}}{\partial x_{1}}&=kx_{1}-2kx_{2}
\end{aligned} ∂ x ˙ 2 ∂ L ∂ x 1 ∂ L = m x ˙ 2 = k x 1 − 2 k x 2 d d t ( ∂ L ∂ x ˙ 2 ) − ∂ L ∂ x 2 = 0 \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \mathscr{L}}{\partial\dot{x}_{2}}\right)-\frac{\partial\mathscr{L}}{\partial x_{2}}=0 d t d ( ∂ x ˙ 2 ∂ L ) − ∂ x 2 ∂ L = 0 得到运动方程
m x ¨ 2 − k x 1 + 2 k x 2 = 0 m\ddot{x}_{2}-kx_{1}+2kx_{2}=0 m x ¨ 2 − k x 1 + 2 k x 2 = 0
x ¨ 2 = k m x 1 − 2 k m x 2 (2.2) \ddot{x}_{2}=\frac{k}{m}x_{1}-\frac{2k}{m}x_{2}\tag{2.2} x ¨ 2 = m k x 1 − m 2 k x 2 ( 2.2 ) (2.1 ) 和 (2.2 ) 与式 (1.8.24) 和 (1.8.25) 相同。
■ ~\tag*{$\blacksquare$} ■ 习题 2.1.3 一个质量为 m m m V ( r , θ , ϕ ) = V ( r ) V(r,\theta,\phi)=V(r) V ( r , θ , ϕ ) = V ( r ) L \mathscr{L} L
解答:
动能和势能分别为
T = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 + r 2 sin 2 θ ϕ ˙ 2 ) V = V ( r ) \begin{aligned}
T&=\frac{1}{2}m(\dot{r}^{2}+r^{2}\dot{\theta}^{2}+r^{2}\sin^{2}\theta\dot{\phi}^{2})\\
V&=V(r)
\end{aligned} T V = 2 1 m ( r ˙ 2 + r 2 θ ˙ 2 + r 2 sin 2 θ ϕ ˙ 2 ) = V ( r ) 于是Lagrange量为
L = T − V = 1 2 m ( r ˙ 2 + r 2 θ ˙ 2 + r 2 sin 2 θ ϕ ˙ 2 ) − V ( r ) \mathscr{L}=T-V=\frac{1}{2}m(\dot{r}^{2}+r^{2}\dot{\theta}^{2}+r^{2}\sin^{2}\theta\dot{\phi}^{2})-V(r) L = T − V = 2 1 m ( r ˙ 2 + r 2 θ ˙ 2 + r 2 sin 2 θ ϕ ˙ 2 ) − V ( r ) ∂ L ∂ r ˙ = m r ˙ , ∂ L ∂ r = m r θ ˙ 2 + m r sin 2 θ ϕ ˙ 2 − ∂ V ( r ) r \frac{\partial \mathscr{L}}{\partial \dot{r}}=m\dot{r},\quad \frac{\partial \mathscr{L}}{\partial r}=mr\dot{\theta}^{2}+mr\sin^{2}\theta\,\dot{\phi}^{2}-\frac{\partial V(r)}{r} ∂ r ˙ ∂ L = m r ˙ , ∂ r ∂ L = m r θ ˙ 2 + m r sin 2 θ ϕ ˙ 2 − r ∂ V ( r ) d d t ( ∂ L ∂ r ˙ ) − ∂ L ∂ r = 0 \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \mathscr{L}}{\partial\dot{r}}\right)-\frac{\partial\mathscr{L}}{\partial r}=0 d t d ( ∂ r ˙ ∂ L ) − ∂ r ∂ L = 0 运动方程为
m r ¨ − m r θ ˙ 2 − m r sin 2 θ ϕ ˙ 2 + ∂ V ( r ) ∂ r = 0 m\ddot{r}-mr\dot{\theta}^{2}-mr\sin^{2}\theta\,\dot{\phi}^{2}+\frac{\partial V(r)}{\partial r}=0 m r ¨ − m r θ ˙ 2 − m r sin 2 θ ϕ ˙ 2 + ∂ r ∂ V ( r ) = 0
θ \theta θ
∂ L ∂ θ ˙ = m r 2 θ ˙ , ∂ θ ∂ r = m r 2 sin θ cos θ ϕ ˙ 2 \frac{\partial \mathscr{L}}{\partial \dot{\theta}}=mr^{2}\dot{\theta},\quad \frac{\partial \mathscr{\theta}}{\partial r}=mr^{2}\sin\theta\cos\theta\,\dot{\phi}^{2} ∂ θ ˙ ∂ L = m r 2 θ ˙ , ∂ r ∂ θ = m r 2 sin θ cos θ ϕ ˙ 2 d d t ( ∂ L ∂ θ ˙ ) − ∂ L ∂ θ = 0 \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \mathscr{L}}{\partial\dot{\theta}}\right)-\frac{\partial\mathscr{L}}{\partial \theta}=0 d t d ( ∂ θ ˙ ∂ L ) − ∂ θ ∂ L = 0 运动方程为
m r 2 θ ¨ + 2 m r r ˙ θ ˙ − m r 2 sin θ cos θ ϕ ˙ 2 = 0 mr^{2}\ddot{\theta}+2mr\dot{r}\dot{\theta}-mr^{2}\sin\theta\cos\theta\,\dot{\phi}^{2}=0 m r 2 θ ¨ + 2 m r r ˙ θ ˙ − m r 2 sin θ cos θ ϕ ˙ 2 = 0
ϕ \phi ϕ
∂ L ∂ ϕ ˙ = m r 2 sin 2 θ ϕ ˙ , ∂ L ∂ ϕ = 0 \frac{\partial \mathscr{L}}{\partial \dot{\phi}}=mr^{2}\sin^{2}\theta\dot{\phi},\quad \frac{\partial \mathscr{L}}{\partial \phi}=0 ∂ ϕ ˙ ∂ L = m r 2 sin 2 θ ϕ ˙ , ∂ ϕ ∂ L = 0 d d t ( ∂ L ∂ r ˙ ) − ∂ L ∂ r = 0 \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \mathscr{L}}{\partial\dot{r}}\right)-\frac{\partial\mathscr{L}}{\partial r}=0 d t d ( ∂ r ˙ ∂ L ) − ∂ r ∂ L = 0 运动方程为
d d t ( m r 2 sin 2 θ ϕ ˙ ) = 0 m r 2 sin 2 θ ϕ ˙ = l ϕ ˙ = l m r 2 sin 2 θ \begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}t}(mr^{2}\sin^{2}\theta\,\dot{\phi})&=0\\
mr^{2}\sin^{2}\theta\,\dot{\phi}&=l\\
\dot{\phi}&=\frac{l}{mr^{2}\sin^{2}\theta}
\end{aligned} d t d ( m r 2 sin 2 θ ϕ ˙ ) m r 2 sin 2 θ ϕ ˙ ϕ ˙ = 0 = l = m r 2 sin 2 θ l 其中 l l l
■ ~\tag*{$\blacksquare$} ■ 2.2 电磁Lagrange量
2.3 二体问题
习题 2.3.1
通过变量变换，由式 (2.3.5) 推导式 (2.3.6)。
解答: 由于
r 1 = r C M + m 2 r m 1 + m 2 r ˙ 1 = r ˙ C M + m 2 r ˙ m 1 + m 2 r 2 = r C M − m 1 r m 1 + m 2 r ˙ 2 = r ˙ C M − m 1 r ˙ m 1 + m 2 \begin{aligned}
\mathbf{r}_1=\mathbf{r}_{\mathrm{CM}}+\frac{m_2 \mathbf{r}}{m_1+m_2}\quad & \quad \dot{\mathbf{r}}_1=\dot{\mathbf{r}}_{\mathrm{CM}}+\frac{m_2 \dot{\mathbf{r}}}{m_1+m_2} \\
\mathbf{r}_2=\mathbf{r}_{\mathrm{CM}}-\frac{m_1 \mathbf{r}}{m_1+m_2}\quad &\quad \dot{\mathbf{r}}_2=\dot{\mathbf{r}}_{\mathrm{CM}}-\frac{m_1 \dot{\mathbf{r}}}{m_1+m_2}
\end{aligned} r 1 = r CM + m 1 + m 2 m 2 r r 2 = r CM − m 1 + m 2 m 1 r r ˙ 1 = r ˙ CM + m 1 + m 2 m 2 r ˙ r ˙ 2 = r ˙ CM − m 1 + m 2 m 1 r ˙ Lagrange量变为
L = 1 2 m 1 ∣ r ˙ 1 ∣ 2 + 1 2 m 2 ∣ r ˙ 2 ∣ 2 − V ( r 1 − r 2 ) = 1 2 m 1 ( r ˙ C M + m 2 r ˙ m 1 + m 2 ) 2 + 1 2 m 2 ( r ˙ C M − m 1 r ˙ m 1 + m 2 ) 2 − V ( r ) = 1 2 m 1 ∣ r ˙ C M ∣ 2 + 1 2 m 1 m 2 2 ( m 1 + m 2 ) 2 ∣ r ˙ ∣ 2 + 1 2 m 2 ∣ r ˙ C M ∣ 2 + 1 2 m 2 m 1 2 ( m 1 + m 2 ) 2 ∣ r ˙ ∣ 2 − V ( r ) = 1 2 ( m 1 + m 2 ) ∣ r ˙ C M ∣ 2 + 1 2 m 1 m 2 ( m 1 + m 2 ) ( m 1 + m 2 ) 2 ∣ r ˙ ∣ 2 − V ( r ) = 1 2 ( m 1 + m 2 ) ∣ r ˙ C M ∣ 2 + 1 2 m 1 m 2 m 1 + m 2 ∣ r ˙ ∣ 2 − V ( r ) \begin{aligned}
\mathscr{L}&=\frac{1}{2} m_{1}\left|\dot{\mathbf{r}}_1\right|^2+\frac{1}{2} m_2\left|\dot{\mathbf{r}}_2\right|^2-V(\mathbf{r}_1-\mathbf{r}_2)\\
&=\frac{1}{2} m_1\left(\dot{\mathbf{r}}_{\mathrm{CM}}+\frac{m_2 \dot{\mathbf{r}}}{m_1+m_2}\right)^{2}+\frac{1}{2}m_{2}\left(\dot{\mathbf{r}}_{\mathrm{CM}}-\frac{m_1 \dot{\mathbf{r}}}{m_1+m_2}\right)^{2}-V(\mathbf{r})\\
&=\frac{1}{2} m_1|\dot{\mathbf{r}}_{\mathrm{CM}}|^{2}+\frac{1}{2}\frac{m_{1}m_2^{2}}{\left(m_1+m_2\right)^{2}}|\dot{\mathbf{r}}|^{2}+\frac{1}{2} m_2|\dot{\mathbf{r}}_{\mathrm{CM}}|^{2}+\frac{1}{2}\frac{m_{2}m_1^{2}}{\left(m_1+m_2\right)^{2}}|\dot{\mathbf{r}}|^{2}-V(\mathbf{r})\\
&=\frac{1}{2}(m_{1}+m_{2})|\dot{\mathbf{r}}_{\mathrm{CM}}|^{2}+\frac{1}{2}\frac{m_{1}m_{2}(m_{1}+m_{2})}{(m_{1}+m_{2})^{2}}|\dot{\mathbf{r}}|^{2}-V(\mathbf{r})\\
&=\frac{1}{2}(m_{1}+m_{2})|\dot{\mathbf{r}}_{\mathrm{CM}}|^{2}+\frac{1}{2}\frac{m_{1}m_{2}}{m_{1}+m_{2}}|\dot{\mathbf{r}}|^{2}-V(\mathbf{r})
\end{aligned} L = 2 1 m 1 ∣ r ˙ 1 ∣ 2 + 2 1 m 2 ∣ r ˙ 2 ∣ 2 − V ( r 1 − r 2 ) = 2 1 m 1 ( r ˙ CM + m 1 + m 2 m 2 r ˙ ) 2 + 2 1 m 2 ( r ˙ CM − m 1 + m 2 m 1 r ˙ ) 2 − V ( r ) = 2 1 m 1 ∣ r ˙ CM ∣ 2 + 2 1 ( m 1 + m 2 ) 2 m 1 m 2 2 ∣ r ˙ ∣ 2 + 2 1 m 2 ∣ r ˙ CM ∣ 2 + 2 1 ( m 1 + m 2 ) 2 m 2 m 1 2 ∣ r ˙ ∣ 2 − V ( r ) = 2 1 ( m 1 + m 2 ) ∣ r ˙ CM ∣ 2 + 2 1 ( m 1 + m 2 ) 2 m 1 m 2 ( m 1 + m 2 ) ∣ r ˙ ∣ 2 − V ( r ) = 2 1 ( m 1 + m 2 ) ∣ r ˙ CM ∣ 2 + 2 1 m 1 + m 2 m 1 m 2 ∣ r ˙ ∣ 2 − V ( r ) ■ ~\tag*{$\blacksquare$} ■ 2.4 粒子有多聪明？
2.5 Hamilton形式
习题 2.5.1
设 T = ∑ i ∑ j T i j ( q ) q ˙ i q ˙ j T=\sum\limits_i \sum\limits_j T_{i j}(q) \dot{q}_i \dot{q}_j T = i ∑ j ∑ T ij ( q ) q ˙ i q ˙ j q ˙ \dot{q} q ˙ ∑ i p i q ˙ i = 2 T \sum\limits_i p_i \dot{q}_i=2 T i ∑ p i q ˙ i = 2 T
解答:
p s = ∂ T ∂ q ˙ s = ∑ i ∑ j T i j ( q ) q ˙ i δ j s + ∑ i ∑ j T i j ( q ) δ i s q ˙ j = ∑ i T i s ( q ) q ˙ i + ∑ j T s j ( q ) q ˙ j \begin{aligned}
p_{s}&=\frac{\partial T}{\partial \dot{q}_{s}}\\
&=\sum_{i}\sum_{j}T_{ij}(q)\dot{q}_{i}\delta_{js}+\sum_{i}\sum_{j}T_{ij}(q)\delta_{is}\dot{q}_{j}\\
&=\sum_{i}T_{is}(q)\dot{q}_{i}+\sum_{j}T_{sj}(q)\dot{q}_{j}
\end{aligned} p s = ∂ q ˙ s ∂ T = i ∑ j ∑ T ij ( q ) q ˙ i δ j s + i ∑ j ∑ T ij ( q ) δ i s q ˙ j = i ∑ T i s ( q ) q ˙ i + j ∑ T s j ( q ) q ˙ j 因此，
∑ s p s q ˙ s = ∑ i T i s ( q ) q ˙ i q ˙ s + ∑ j T s j ( q ) q ˙ j q ˙ s = T + T = 2 T \begin{aligned}
\sum_{s}p_{s}\dot{q}_{s}&=\sum_{i}T_{is}(q)\dot{q}_{i}\dot{q}_{s}+\sum_{j}T_{sj}(q)\dot{q}_{j}\dot{q}_{s}\\
&=T+T\\
&=2T
\end{aligned} s ∑ p s q ˙ s = i ∑ T i s ( q ) q ˙ i q ˙ s + j ∑ T s j ( q ) q ˙ j q ˙ s = T + T = 2 T ■ ~\tag*{$\blacksquare$} ■ 习题 2.5.2 利用能量守恒，证明谐振子在相空间中的轨迹是形如 ( x / a ) 2 + ( p / b ) 2 = 1 (x / a)^2+(p / b)^2=1 ( x / a ) 2 + ( p / b ) 2 = 1 a 2 = 2 E / k a^2=2 E / k a 2 = 2 E / k b 2 = 2 m E b^2=2 m E b 2 = 2 m E
解答:
Lagrange量为
L = 1 2 m x ˙ 2 − 1 2 k x 2 \mathscr{L}=\frac{1}{2}m\dot{x}^{2}-\frac{1}{2}kx^{2} L = 2 1 m x ˙ 2 − 2 1 k x 2 因此动量为
p = ∂ L ∂ x ˙ = m x ˙ p=\frac{\partial \mathscr{L}}{\partial \dot{x}}=m\dot{x} p = ∂ x ˙ ∂ L = m x ˙ Hamilton量为
H = p x ˙ − L = p 2 2 m + 1 2 k x 2 \mathscr{H}=p\dot{x}-\mathscr{L}=\frac{p^{2}}{2m}+\frac{1}{2}kx^{2} H = p x ˙ − L = 2 m p 2 + 2 1 k x 2 由于 L \mathscr{L} L t t t H \mathscr{H} H H = E \mathscr{H}=E H = E E E E
1 2 k x 2 + p 2 2 m = E \frac{1}{2}kx^{2}+\frac{p^{2}}{2m}=E 2 1 k x 2 + 2 m p 2 = E 若记 a 2 = 2 E / k a^{2}=2E/k a 2 = 2 E / k b 2 = 2 m E b^{2}=2mE b 2 = 2 m E
( x a ) 2 + ( p b ) 2 = 1 \left(\frac{x}{a}\right)^{2}+\left(\frac{p}{b}\right)^{2}=1 ( a x ) 2 + ( b p ) 2 = 1 ■ ~\tag*{$\blacksquare$} ■ 习题 2.5.3 使用Hamilton形式求解习题 2.1.2。
解答:
从系统的Lagrange量出发
L = 1 2 m ( x ˙ 1 2 + x ˙ 2 2 ) − k ( x 1 2 + x 2 2 − x 1 x 2 ) \mathscr{L}=\frac{1}{2}m(\dot{x}_{1}^{2}+\dot{x}_{2}^{2})-k(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}) L = 2 1 m ( x ˙ 1 2 + x ˙ 2 2 ) − k ( x 1 2 + x 2 2 − x 1 x 2 ) 于是动量为
p 1 = ∂ L ∂ x ˙ 1 = m x ˙ 1 ⇒ x ˙ 1 = p 1 m p 2 = ∂ L ∂ x ˙ 2 = m x ˙ 2 ⇒ x ˙ 2 = p 2 m \begin{aligned}
p_{1}&=\frac{\partial \mathscr{L}}{\partial \dot{x}_{1}}=m\dot{x}_{1}\quad \Rightarrow \quad \dot{x}_{1}=\frac{p_{1}}{m}\\
p_{2}&=\frac{\partial \mathscr{L}}{\partial \dot{x}_{2}}=m\dot{x}_{2}\quad \Rightarrow \quad \dot{x}_{2}=\frac{p_{2}}{m}
\end{aligned} p 1 p 2 = ∂ x ˙ 1 ∂ L = m x ˙ 1 ⇒ x ˙ 1 = m p 1 = ∂ x ˙ 2 ∂ L = m x ˙ 2 ⇒ x ˙ 2 = m p 2 于是系统的Hamilton量为
H = p 1 x ˙ 1 + p 2 x ˙ 2 − L = p 1 2 m + p 2 2 m − p 1 2 2 m − p 2 2 2 m + k ( x 1 2 + x 2 2 − x 1 x 2 ) = p 1 2 2 m + p 2 2 2 m + k ( x 1 2 + x 2 2 − x 1 x 2 ) \begin{aligned}
\mathscr{H}&=p_{1}\dot{x}_{1}+p_{2}\dot{x}_{2}-\mathscr{L}\\
&=\frac{p_{1}^{2}}{m}+\frac{p_{2}^{2}}{m}-\frac{p_{1}^{2}}{2m}-\frac{p_{2}^{2}}{2m}+k(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2})\\
&=\frac{p_{1}^{2}}{2m}+\frac{p_{2}^{2}}{2m}+k(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2})
\end{aligned} H = p 1 x ˙ 1 + p 2 x ˙ 2 − L = m p 1 2 + m p 2 2 − 2 m p 1 2 − 2 m p 2 2 + k ( x 1 2 + x 2 2 − x 1 x 2 ) = 2 m p 1 2 + 2 m p 2 2 + k ( x 1 2 + x 2 2 − x 1 x 2 ) { x ˙ 1 = ∂ H ∂ p 1 = p 1 m p ˙ 1 = − ∂ H ∂ x 1 = − 2 k x 1 + k x 2 \left\{\begin{aligned}
\dot{x}_{1}&=\frac{\partial\mathscr{H}}{\partial p_{1}}=\frac{p_{1}}{m}\\
\dot{p}_{1}&=-\frac{\partial\mathscr{H}}{\partial x_{1}}=-2kx_{1}+kx_{2}
\end{aligned}\right. ⎩ ⎨ ⎧ x ˙ 1 p ˙ 1 = ∂ p 1 ∂ H = m p 1 = − ∂ x 1 ∂ H = − 2 k x 1 + k x 2 由第一个方程可知 p 1 = m x ˙ 1 p_{1}=m\dot{x}_{1} p 1 = m x ˙ 1 p ˙ 1 = m x ¨ 1 \dot{p}_{1}=m\ddot{x}_{1} p ˙ 1 = m x ¨ 1
m x ¨ 1 = − 2 k x 1 + k x 2 x ¨ 1 = − 2 k m x 1 + k m x 2 \begin{aligned}
m\ddot{x}_{1}&=-2kx_{1}+kx_{2}\\
\ddot{x}_{1}&=-\frac{2k}{m}x_{1}+\frac{k}{m}x_{2}
\end{aligned} m x ¨ 1 x ¨ 1 = − 2 k x 1 + k x 2 = − m 2 k x 1 + m k x 2 { x ˙ 2 = ∂ H ∂ p 2 = p 2 m p ˙ 2 = − ∂ H ∂ x 2 = − 2 k x 2 + k x 1 \left\{\begin{aligned}
\dot{x}_{2}&=\frac{\partial\mathscr{H}}{\partial p_{2}}=\frac{p_{2}}{m}\\
\dot{p}_{2}&=-\frac{\partial\mathscr{H}}{\partial x_{2}}=-2kx_{2}+kx_{1}
\end{aligned}\right. ⎩ ⎨ ⎧ x ˙ 2 p ˙ 2 = ∂ p 2 ∂ H = m p 2 = − ∂ x 2 ∂ H = − 2 k x 2 + k x 1 由第一个方程可知 p 2 = m x ˙ 2 p_{2}=m\dot{x}_{2} p 2 = m x ˙ 2 p ˙ 2 = m x ¨ 2 \dot{p}_{2}=m\ddot{x}_{2} p ˙ 2 = m x ¨ 2
m x ¨ 2 = − 2 k x 2 + k x 1 x ¨ 2 = k m x 1 − 2 k m x 2 \begin{aligned}
m\ddot{x}_{2}&=-2kx_{2}+kx_{1}\\
\ddot{x}_{2}&=\frac{k}{m}x_{1}-\frac{2k}{m}x_{2}
\end{aligned} m x ¨ 2 x ¨ 2 = − 2 k x 2 + k x 1 = m k x 1 − m 2 k x 2 ■ ~\tag*{$\blacksquare$} ■ 习题 2.5.4
证明与式 (2.3.6) 中的 L \mathscr{L} L H = ∣ p C M ∣ 2 / 2 M + ∣ p ∣ 2 / \mathscr{H}=\left|\mathbf{p}_{\mathrm{CM}}\right|^2 / 2 M+|\mathbf{p}|^2 / H = ∣ p CM ∣ 2 /2 M + ∣ p ∣ 2 / 2 μ + V ( r ) 2 \mu+V(\mathbf{r}) 2 μ + V ( r ) M M M μ \mu μ p C M \mathbf{p}_{\mathrm{CM}} p CM p \mathbf{p} p r C M \mathbf{r}_{\mathrm{CM}} r CM r \mathbf{r} r
解答:
从Lagrange量出发
L = 1 2 ( m 1 + m 2 ) ∣ r ˙ C M ∣ 2 + 1 2 m 1 m 2 m 1 + m 2 ∣ r ˙ ∣ 2 − V ( r ) = 1 2 M ∣ r ˙ C M ∣ 2 + 1 2 μ ∣ r ˙ ∣ 2 − V ( r ) \begin{aligned}
\mathscr{L}&=\frac{1}{2}(m_{1}+m_{2})|\dot{\mathbf{r}}_{\mathrm{CM}}|^{2}+\frac{1}{2}\frac{m_{1}m_{2}}{m_{1}+m_{2}}|\dot{\mathbf{r}}|^{2}-V(\mathbf{r})\\
&=\frac{1}{2}M|\dot{\mathbf{r}}_{\mathrm{CM}}|^{2}+\frac{1}{2}\mu|\dot{\mathbf{r}}|^{2}-V(\mathbf{r})
\end{aligned} L = 2 1 ( m 1 + m 2 ) ∣ r ˙ CM ∣ 2 + 2 1 m 1 + m 2 m 1 m 2 ∣ r ˙ ∣ 2 − V ( r ) = 2 1 M ∣ r ˙ CM ∣ 2 + 2 1 μ ∣ r ˙ ∣ 2 − V ( r ) 其中总质量 M = m 1 + m 2 M=m_{1}+m_{2} M = m 1 + m 2 μ = m 1 m 2 m 1 + m 2 \mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}} μ = m 1 + m 2 m 1 m 2
∣ p C M ∣ = ∂ L ∂ ∣ r ˙ C M ∣ = M ∣ r ˙ C M ∣ ⇒ ∣ r ˙ C M ∣ = ∣ p C M ∣ M ∣ p ∣ = ∂ L ∂ ∣ r ˙ ∣ = μ ∣ r ˙ ∣ ⇒ ∣ r ˙ ∣ = ∣ p ∣ μ \begin{aligned}
|\mathbf{p}_{\mathrm{CM}}|&=\frac{\partial \mathscr{L}}{\partial |\dot{\mathbf{r}}_{\mathrm{CM}}|}=M|\dot{\mathbf{r}}_{\mathrm{CM}}|\quad &\Rightarrow& \quad |\dot{\mathbf{r}}_{\mathrm{CM}}|=\frac{|\mathbf{p}_{\mathrm{CM}}|}{M}\\
|\mathbf{p}|&=\frac{\partial \mathscr{L}}{\partial |\dot{\mathbf{r}}|}=\mu|\dot{\mathbf{r}}|\quad &\Rightarrow& \quad |\dot{\mathbf{r}}|=\frac{|\mathbf{p}|}{\mu}
\end{aligned} ∣ p CM ∣ ∣ p ∣ = ∂ ∣ r ˙ CM ∣ ∂ L = M ∣ r ˙ CM ∣ = ∂ ∣ r ˙ ∣ ∂ L = μ ∣ r ˙ ∣ ⇒ ⇒ ∣ r ˙ CM ∣ = M ∣ p CM ∣ ∣ r ˙ ∣ = μ ∣ p ∣ 因此Hamilton量为
H = p C M ⋅ r ˙ C M + p ⋅ r ˙ − L = ∣ p C M ∣ ∣ r ˙ C M ∣ + ∣ p ∣ ∣ r ˙ ∣ − 1 2 M ∣ r ˙ C M ∣ 2 − 1 2 μ ∣ r ˙ ∣ 2 + V ( r ) = ∣ p C M ∣ 2 M + ∣ p ∣ 2 μ − 1 2 M ∣ p C M ∣ 2 M 2 − 1 2 μ ∣ p ∣ 2 μ 2 + V ( r ) = ∣ p C M ∣ 2 2 M + ∣ p ∣ 2 2 μ + V ( r ) \begin{aligned}
\mathscr{H}&=\mathbf{p}_{\mathrm{CM}}\cdot \dot{\mathbf{r}}_{\mathrm{CM}}+\mathbf{p}\cdot\dot{\mathbf{r}}-\mathscr{L}\\
&=|\mathbf{p}_{\mathrm{CM}}||\dot{\mathbf{r}}_{\mathrm{CM}}|+|\mathbf{p}||\dot{\mathbf{r}}|-\frac{1}{2}M|\dot{\mathbf{r}}_{\mathrm{CM}}|^{2}-\frac{1}{2}\mu|\dot{\mathbf{r}}|^{2}+V(\mathbf{r})\\
&=\frac{|\mathbf{p}_{\mathrm{CM}}|^{2}}{M}+\frac{|\mathbf{p}|^{2}}{\mu}-\frac{1}{2}M\frac{|\mathbf{p}_{\mathrm{CM}}|^{2}}{M^{2}}-\frac{1}{2}\mu\frac{|\mathbf{p}|^{2}}{\mu^{2}}+V(\mathbf{r})\\
&=\frac{|\mathbf{p}_{\mathrm{CM}}|^{2}}{2M}+\frac{|\mathbf{p}|^{2}}{2\mu}+V(\mathbf{r})
\end{aligned} H = p CM ⋅ r ˙ CM + p ⋅ r ˙ − L = ∣ p CM ∣∣ r ˙ CM ∣ + ∣ p ∣∣ r ˙ ∣ − 2 1 M ∣ r ˙ CM ∣ 2 − 2 1 μ ∣ r ˙ ∣ 2 + V ( r ) = M ∣ p CM ∣ 2 + μ ∣ p ∣ 2 − 2 1 M M 2 ∣ p CM ∣ 2 − 2 1 μ μ 2 ∣ p ∣ 2 + V ( r ) = 2 M ∣ p CM ∣ 2 + 2 μ ∣ p ∣ 2 + V ( r ) ■ ~\tag*{$\blacksquare$} ■ 2.6 Hamilton框架中的电磁力
2.7 循环坐标、Poisson括号与正则变换
习题 2.7.1
证明
{ ω , λ } = − { λ , ω } { ω , λ + σ } = { ω , λ } + { ω , σ } { ω , λ σ } = { ω , λ } σ + λ { ω , σ } \begin{gathered}
\{\omega, \lambda\}=-\{\lambda, \omega\} \\
\{\omega, \lambda+\sigma\}=\{\omega, \lambda\}+\{\omega, \sigma\} \\
\{\omega, \lambda \sigma\}=\{\omega, \lambda\} \sigma+\lambda\{\omega, \sigma\}
\end{gathered} { ω , λ } = − { λ , ω } { ω , λ + σ } = { ω , λ } + { ω , σ } { ω , λσ } = { ω , λ } σ + λ { ω , σ } 注意，上述关系与对易子的式 (1.5.10) 和 (1.5.11) 十分相似。
解答:
{ ω , λ } = ∑ i ( ∂ ω ∂ q i ∂ λ ∂ p i − ∂ ω ∂ p i ∂ λ ∂ q i ) = − ∑ i ( ∂ λ ∂ q i ∂ ω ∂ p i − ∂ λ ∂ p i ∂ ω ∂ q i ) = − { λ , ω } \{\omega, \lambda\}=\sum_i\left(\frac{\partial \omega}{\partial q_i} \frac{\partial \lambda}{\partial p_i}-\frac{\partial \omega}{\partial p_i} \frac{\partial \lambda}{\partial q_i}\right)=-\sum_i\left(\frac{\partial \lambda}{\partial q_i} \frac{\partial \omega}{\partial p_i}-\frac{\partial \lambda}{\partial p_i} \frac{\partial \omega}{\partial q_i}\right)=-\{\lambda, \omega\} { ω , λ } = i ∑ ( ∂ q i ∂ ω ∂ p i ∂ λ − ∂ p i ∂ ω ∂ q i ∂ λ ) = − i ∑ ( ∂ q i ∂ λ ∂ p i ∂ ω − ∂ p i ∂ λ ∂ q i ∂ ω ) = − { λ , ω } { ω , λ + σ } = ∑ i ( ∂ ω ∂ q i ⋅ ∂ ( λ + σ ) ∂ p i − ∂ ω ∂ p i ⋅ ∂ ( λ + σ ) ∂ q i ) = ∑ i [ ∂ ω ∂ q i ⋅ ( ∂ λ ∂ p i + ∂ σ ∂ p i ) − ∂ ω ∂ p i ⋅ ( ∂ λ ∂ q i + ∂ σ ∂ q i ) ] = ∑ i [ ( ∂ ω ∂ q i ∂ λ ∂ p i − ∂ ω ∂ p i ∂ λ ∂ q i ) + ( ∂ ω ∂ q i ∂ σ ∂ p i − ∂ ω ∂ p i ∂ σ ∂ q i ) ] = ∑ i ( ∂ ω ∂ q i ∂ λ ∂ p i − ∂ ω ∂ p i ∂ λ ∂ q i ) + ∑ i ( ∂ ω ∂ q i ∂ σ ∂ p i − ∂ ω ∂ p i ∂ σ ∂ q i ) = { ω , λ } + { ω , σ } \begin{aligned}
\{\omega, \lambda+\sigma\} & =\sum_i\left(\frac{\partial \omega}{\partial q_i} \cdot \frac{\partial(\lambda+\sigma)}{\partial p_i}-\frac{\partial \omega}{\partial p_i} \cdot \frac{\partial(\lambda+\sigma)}{\partial q_i}\right) \\
& =\sum_i\left[\frac{\partial \omega}{\partial q_i} \cdot\left(\frac{\partial \lambda}{\partial p_i}+\frac{\partial \sigma}{\partial p_i}\right)-\frac{\partial \omega}{\partial p_i} \cdot\left(\frac{\partial \lambda}{\partial q_i}+\frac{\partial \sigma}{\partial q_i}\right)\right] \\
& =\sum_i\left[\left(\frac{\partial \omega}{\partial q_i} \frac{\partial \lambda}{\partial p_i}-\frac{\partial \omega}{\partial p_i} \frac{\partial \lambda}{\partial q_i}\right)+\left(\frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial p_i}-\frac{\partial \omega}{\partial p_i} \frac{\partial \sigma}{\partial q_i}\right)\right] \\
& =\sum_i\left(\frac{\partial \omega}{\partial q_i} \frac{\partial \lambda}{\partial p_i}-\frac{\partial \omega}{\partial p_i}\frac{\partial \lambda}{\partial q_{i}}\right)+\sum_i\left(\frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial p_i}-\frac{\partial \omega}{\partial p_i}\frac{\partial \sigma}{\partial q_{i}}\right) \\
& =\{\omega, \lambda\}+\{\omega, \sigma\}
\end{aligned} { ω , λ + σ } = i ∑ ( ∂ q i ∂ ω ⋅ ∂ p i ∂ ( λ + σ ) − ∂ p i ∂ ω ⋅ ∂ q i ∂ ( λ + σ ) ) = i ∑ [ ∂ q i ∂ ω ⋅ ( ∂ p i ∂ λ + ∂ p i ∂ σ ) − ∂ p i ∂ ω ⋅ ( ∂ q i ∂ λ + ∂ q i ∂ σ ) ] = i ∑ [ ( ∂ q i ∂ ω ∂ p i ∂ λ − ∂ p i ∂ ω ∂ q i ∂ λ ) + ( ∂ q i ∂ ω ∂ p i ∂ σ − ∂ p i ∂ ω ∂ q i ∂ σ ) ] = i ∑ ( ∂ q i ∂ ω ∂ p i ∂ λ − ∂ p i ∂ ω ∂ q i ∂ λ ) + i ∑ ( ∂ q i ∂ ω ∂ p i ∂ σ − ∂ p i ∂ ω ∂ q i ∂ σ ) = { ω , λ } + { ω , σ } { ω , λ σ } = ∑ i [ ∂ ω ∂ q i ∂ ( λ σ ) ∂ p i − ∂ ω ∂ p i ⋅ ∂ ( λ σ ) ∂ q i ] = ∑ i [ λ ∂ ω ∂ q i ∂ σ ∂ p i + ∂ ω ∂ q i ∂ λ ∂ p i σ − λ ∂ ω ∂ p i ∂ σ ∂ q i − ∂ ω ∂ p i ∂ λ ∂ q i σ ] = λ ∑ i ( ∂ ω ∂ q i ∂ σ ∂ p i − ∂ ω ∂ p i ∂ σ ∂ q i ) + ∑ i ( ∂ ω ∂ q i ∂ λ ∂ p i − ∂ ω ∂ p i ∂ λ ∂ q i ) σ = λ { ω , σ } + { ω , λ } σ . \begin{aligned}
\{\omega, \lambda \sigma\} & =\sum_i\left[\frac{\partial \omega}{\partial q_i} \frac{\partial(\lambda \sigma)}{\partial p_i}-\frac{\partial \omega}{\partial p_i} \cdot \frac{\partial(\lambda \sigma)}{\partial q_i}\right] \\
& =\sum_i\left[\lambda \frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial p_i}+\frac{\partial \omega}{\partial q_i} \frac{\partial \lambda}{\partial p_i}\sigma-\lambda \frac{\partial \omega}{\partial p_i} \frac{\partial \sigma}{\partial q_i}-\frac{\partial \omega}{\partial p_i} \frac{\partial \lambda}{\partial q_i} \sigma\right] \\
& =\lambda \sum_i\left(\frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial p_i}-\frac{\partial \omega}{\partial p_i} \frac{\partial \sigma}{\partial q_i}\right)+\sum_i\left(\frac{\partial \omega}{\partial q_i} \frac{\partial \lambda}{\partial p_i}-\frac{\partial \omega}{\partial p_i} \frac{\partial \lambda}{\partial q_i}\right) \sigma \\
& =\lambda\{\omega, \sigma\}+\{\omega, \lambda\} \sigma .
\end{aligned} { ω , λσ } = i ∑ [ ∂ q i ∂ ω ∂ p i ∂ ( λσ ) − ∂ p i ∂ ω ⋅ ∂ q i ∂ ( λσ ) ] = i ∑ [ λ ∂ q i ∂ ω ∂ p i ∂ σ + ∂ q i ∂ ω ∂ p i ∂ λ σ − λ ∂ p i ∂ ω ∂ q i ∂ σ − ∂ p i ∂ ω ∂ q i ∂ λ σ ] = λ i ∑ ( ∂ q i ∂ ω ∂ p i ∂ σ − ∂ p i ∂ ω ∂ q i ∂ σ ) + i ∑ ( ∂ q i ∂ ω ∂ p i ∂ λ − ∂ p i ∂ ω ∂ q i ∂ λ ) σ = λ { ω , σ } + { ω , λ } σ . ■ ~\tag*{$\blacksquare$} ■ 习题 2.7.2
(i) 验证式 (2.7.4) 和 (2.7.5)。(ii) 考虑二维问题 H = p x 2 + p y 2 + a x 2 + b y 2 \mathscr{H}=p_x^2+p_y^2+a x^2+b y^2 H = p x 2 + p y 2 + a x 2 + b y 2 a = b a=b a = b { l z , H } \left\{l_z, \mathscr{H}\right\} { l z , H }
解答:
(i)
{ q i , q j } : = ∑ k ( ∂ q i ∂ q k ⋅ ∂ q j ∂ p k − ∂ q i ∂ p k ⋅ ∂ q j ∂ q k ) = ∑ k ( ∂ q i ∂ q k ⋅ 0 − 0 ⋅ ∂ q j ∂ q k ) = 0 { p i , p j } : = ∑ k ( ∂ p i ∂ q k ⋅ ∂ p j ∂ p k − ∂ p i ∂ p k ⋅ ∂ p j ∂ q k ) = ∑ k ( 0 ⋅ ∂ p j ∂ p k − ∂ p i ∂ p k ⋅ 0 ) = 0 { q i , p j } : = ∑ k ( ∂ q i ∂ q k ⋅ ∂ p j ∂ p k − ∂ q i ∂ p k ⋅ ∂ p j ∂ q k ) = ∑ k ( δ i k δ j k − 0 ⋅ 0 ) = δ i j \begin{aligned}
\left\{q_i, q_j\right\} & :=\sum_k\left(\frac{\partial q_i}{\partial q_k} \cdot \frac{\partial q_j}{\partial p_k}-\frac{\partial q_i}{\partial p_k} \cdot \frac{\partial q_j}{\partial q_k}\right)=\sum_k\left(\frac{\partial q_i}{\partial q_k} \cdot 0-0 \cdot \frac{\partial q_j}{\partial q_k}\right)=0 \\
\left\{p_i, p_j\right\} & :=\sum_k\left(\frac{\partial p_i}{\partial q_k} \cdot \frac{\partial p_j}{\partial p_k}-\frac{\partial p_i}{\partial p_k} \cdot \frac{\partial p_j}{\partial q_k}\right)=\sum_k\left(0 \cdot \frac{\partial p_j}{\partial p_k}-\frac{\partial p_i}{\partial p_k} \cdot 0\right)=0 \\
\left\{q_i, p_j\right\} & :=\sum_k\left(\frac{\partial q_i}{\partial q_k} \cdot \frac{\partial p_j}{\partial p_k}-\frac{\partial q_i}{\partial p_k} \cdot \frac{\partial p_j}{\partial q_k}\right)=\sum_k\left(\delta_{i k} \delta_{j k}-0 \cdot 0\right)=\delta_{i j}
\end{aligned} { q i , q j } { p i , p j } { q i , p j } := k ∑ ( ∂ q k ∂ q i ⋅ ∂ p k ∂ q j − ∂ p k ∂ q i ⋅ ∂ q k ∂ q j ) = k ∑ ( ∂ q k ∂ q i ⋅ 0 − 0 ⋅ ∂ q k ∂ q j ) = 0 := k ∑ ( ∂ q k ∂ p i ⋅ ∂ p k ∂ p j − ∂ p k ∂ p i ⋅ ∂ q k ∂ p j ) = k ∑ ( 0 ⋅ ∂ p k ∂ p j − ∂ p k ∂ p i ⋅ 0 ) = 0 := k ∑ ( ∂ q k ∂ q i ⋅ ∂ p k ∂ p j − ∂ p k ∂ q i ⋅ ∂ q k ∂ p j ) = k ∑ ( δ ik δ j k − 0 ⋅ 0 ) = δ ij 以及
{ q i , H } : = ∑ k ( ∂ q i ∂ q k ⋅ ∂ H ∂ p k − ∂ q i ∂ p k ⋅ ∂ H ∂ q k ) = ∑ k ( δ i k ⋅ ∂ H ∂ p k − 0 ⋅ ∂ H ∂ q k ) = ∂ H ∂ p i = q ˙ i { p i , H } : = ∑ k ( ∂ p i ∂ q k ⋅ ∂ H ∂ p k − ∂ p i ∂ p k ⋅ ∂ H ∂ q k ) = ∑ k ( 0 ⋅ ∂ H ∂ p k − δ i k ⋅ ∂ H ∂ q k ) = − ∂ H ∂ q i = p ˙ i \begin{aligned}
\left\{q_i, \mathscr{H}\right\} & :=\sum_k\left(\frac{\partial q_i}{\partial q_k} \cdot \frac{\partial \mathscr{H}}{\partial p_k}-\frac{\partial q_i}{\partial p_k} \cdot \frac{\partial \mathscr{H}}{\partial q_k}\right)=\sum_k\left(\delta_{i k} \cdot \frac{\partial \mathscr{H}}{\partial p_k}-0 \cdot \frac{\partial \mathscr{H}}{\partial q_k}\right)\\
&=\frac{\partial \mathscr{H}}{\partial p_i}=\dot{q}_i \\
\left\{p_i, \mathscr{H}\right\} & :=\sum_k\left(\frac{\partial p_i}{\partial q_k} \cdot \frac{\partial \mathscr{H}}{\partial p_k}-\frac{\partial p_i}{\partial p_k} \cdot \frac{\partial \mathscr{H}}{\partial q_k}\right)=\sum_k\left(0 \cdot \frac{\partial \mathscr{H}}{\partial p_k}-\delta_{i k} \cdot \frac{\partial \mathscr{H}}{\partial q_k}\right)\\
&=-\frac{\partial \mathscr{H}}{\partial q_i}=\dot{p}_i
\end{aligned} { q i , H } { p i , H } := k ∑ ( ∂ q k ∂ q i ⋅ ∂ p k ∂ H − ∂ p k ∂ q i ⋅ ∂ q k ∂ H ) = k ∑ ( δ ik ⋅ ∂ p k ∂ H − 0 ⋅ ∂ q k ∂ H ) = ∂ p i ∂ H = q ˙ i := k ∑ ( ∂ q k ∂ p i ⋅ ∂ p k ∂ H − ∂ p k ∂ p i ⋅ ∂ q k ∂ H ) = k ∑ ( 0 ⋅ ∂ p k ∂ H − δ ik ⋅ ∂ q k ∂ H ) = − ∂ q i ∂ H = p ˙ i (ii) 给定的Hamilton量为 H = p x 2 + p y 2 + a x 2 + b y 2 \mathscr{H}=p_{x}^{2}+p_{y}^{2}+ax^{2}+by^{2} H = p x 2 + p y 2 + a x 2 + b y 2 a = b a=b a = b H \mathscr{H} H x − y x-y x − y p x − p y p_{x}-p_{y} p x − p y l z l_{z} l z { l z , H } = 0 \{l_{z},\mathscr{H}\}=0 { l z , H } = 0
{ l z , H } = ∑ k ( ∂ l z ∂ q k ⋅ ∂ H ∂ p k − ∂ l z ∂ p k ⋅ ∂ H ∂ q k ) = ∂ l z ∂ x ⋅ ∂ H ∂ p x + ∂ l z ∂ y ⋅ ∂ H ∂ p y − ∂ l z ∂ p x ⋅ ∂ H ∂ x − ∂ l z ∂ p y ⋅ ∂ H ∂ y \begin{aligned}
\left\{l_z, \mathscr{H}\right\}&=\sum_k\left(\frac{\partial l_z}{\partial q_k} \cdot \frac{\partial \mathscr{H}}{\partial p_k}-\frac{\partial l_z}{\partial p_k} \cdot \frac{\partial \mathscr{H}}{\partial q_k}\right)\\
&=\frac{\partial l_z}{\partial x} \cdot \frac{\partial \mathscr{H}}{\partial p_x}+\frac{\partial l_z}{\partial y} \cdot \frac{\partial \mathscr{H}}{\partial p_y}-\frac{\partial l_z}{\partial p_x} \cdot \frac{\partial \mathscr{H}}{\partial x}-\frac{\partial l_z}{\partial p_y} \cdot \frac{\partial \mathscr{H}}{\partial y}
\end{aligned} { l z , H } = k ∑ ( ∂ q k ∂ l z ⋅ ∂ p k ∂ H − ∂ p k ∂ l z ⋅ ∂ q k ∂ H ) = ∂ x ∂ l z ⋅ ∂ p x ∂ H + ∂ y ∂ l z ⋅ ∂ p y ∂ H − ∂ p x ∂ l z ⋅ ∂ x ∂ H − ∂ p y ∂ l z ⋅ ∂ y ∂ H 但是
∂ H ∂ p k = 2 p k , ∂ l z ∂ p k = ∂ ( x p y − y p x ) ∂ p k = ( ∂ l z ∂ p x , ∂ l z ∂ p y ) = ( − y , x ) , ∂ H ∂ x k = ( ∂ H ∂ x , ∂ H ∂ y ) = ( 2 a x , 2 b y ) , ∂ l z ∂ q k = ( ∂ l z ∂ x , ∂ l z ∂ y ) = ( p y , − p x ) \begin{aligned}
\frac{\partial \mathscr{H}}{\partial p_k}&=2 p_k, \quad \frac{\partial l_z}{\partial p_k}=\frac{\partial\left(x p_y-y p_x\right)}{\partial p_k}=\left(\frac{\partial l_z}{\partial p_x}, \frac{\partial l_z}{\partial p_y}\right)=(-y, x), \\
\frac{\partial \mathscr{H}}{\partial x_k}&=\left(\frac{\partial \mathscr{H}}{\partial x}, \frac{\partial \mathscr{H}}{\partial y}\right)=(2 a x, 2 b y),\quad \frac{\partial l_z}{\partial q_k}=\left(\frac{\partial l_z}{\partial x}, \frac{\partial l_z}{\partial y}\right)=\left(p_y,-p_x\right)
\end{aligned} ∂ p k ∂ H ∂ x k ∂ H = 2 p k , ∂ p k ∂ l z = ∂ p k ∂ ( x p y − y p x ) = ( ∂ p x ∂ l z , ∂ p y ∂ l z ) = ( − y , x ) , = ( ∂ x ∂ H , ∂ y ∂ H ) = ( 2 a x , 2 b y ) , ∂ q k ∂ l z = ( ∂ x ∂ l z , ∂ y ∂ l z ) = ( p y , − p x ) 因此
{ l z , H } = p y ⋅ 2 p x + ( − p x ) ⋅ 2 p y − ( − y ) ⋅ 2 a x − x ⋅ 2 b y = 2 x y ( a − b ) \left\{l_z, \mathscr{H}\right\}=p_y \cdot 2 p_x+\left(-p_x\right) \cdot 2 p_y-(-y) \cdot 2 a x-x \cdot 2 b y=2 x y(a-b) { l z , H } = p y ⋅ 2 p x + ( − p x ) ⋅ 2 p y − ( − y ) ⋅ 2 a x − x ⋅ 2 b y = 2 x y ( a − b ) 当 a = b a=b a = b
■ ~\tag*{$\blacksquare$} ■ 习题 2.7.3
从式 (2.7.14) 出发，补全推导式 (2.7.18) 所缺少的步骤。
解答:
考虑如下变换：
q ˉ i = q ˉ i ( q , p ) p ˉ i = p ˉ i ( q , p ) \begin{aligned}
\bar{q}_{i}&=\bar{q}_{i}(q,p)\\
\bar{p}_{i}&=\bar{p}_{i}(q,p)
\end{aligned} q ˉ i p ˉ i = q ˉ i ( q , p ) = p ˉ i ( q , p ) 如果该变换是正则变换，那么变量 q ˉ i \bar{q}_{i} q ˉ i p ˉ i \bar{p}_{i} p ˉ i
q ˉ ˙ i = ∂ H ∂ p ˉ i p ˉ ˙ i = − ∂ H ∂ q ˉ i \begin{aligned}
\dot{\bar{q}}_{i}&=\frac{\partial\mathscr{H}}{\partial \bar{p}_{i}}\\
\dot{\bar{p}}_{i}&=-\frac{\partial\mathscr{H}}{\partial \bar{q}_{i}}
\end{aligned} q ˉ ˙ i p ˉ ˙ i = ∂ p ˉ i ∂ H = − ∂ q ˉ i ∂ H 若将Hamilton量 H \mathscr{H} H
∂ H ( q ˉ , p ˉ ) ∂ p i = ∑ k ( ∂ H ∂ q ˉ k ∂ q ˉ k ∂ p i + ∂ H ∂ p ˉ k ∂ p ˉ k ∂ p i ) ∂ H ( q ˉ , p ˉ ) ∂ q i = ∑ k ( ∂ H ∂ q ˉ k ∂ q ˉ k ∂ q i + ∂ H ∂ p ˉ k ∂ p ˉ k ∂ q i ) \begin{aligned}
\frac{\partial \mathscr{H}(\bar{q},\bar{p})}{\partial p_{i}}&=\sum_{k}\left(\frac{\partial \mathscr{H}}{\partial \bar{q}_{k}}\frac{\partial \bar{q}_{k}}{\partial p_{i}}+\frac{\partial \mathscr{H}}{\partial \bar{p}_{k}}\frac{\partial \bar{p}_{k}}{\partial p_{i}}\right)\\
\frac{\partial \mathscr{H}(\bar{q},\bar{p})}{\partial q_{i}}&=\sum_{k}\left(\frac{\partial \mathscr{H}}{\partial \bar{q}_{k}}\frac{\partial \bar{q}_{k}}{\partial q_{i}}+\frac{\partial \mathscr{H}}{\partial \bar{p}_{k}}\frac{\partial \bar{p}_{k}}{\partial q_{i}}\right)
\end{aligned} ∂ p i ∂ H ( q ˉ , p ˉ ) ∂ q i ∂ H ( q ˉ , p ˉ ) = k ∑ ( ∂ q ˉ k ∂ H ∂ p i ∂ q ˉ k + ∂ p ˉ k ∂ H ∂ p i ∂ p ˉ k ) = k ∑ ( ∂ q ˉ k ∂ H ∂ q i ∂ q ˉ k + ∂ p ˉ k ∂ H ∂ q i ∂ p ˉ k ) 任意函数 ω \omega ω H \mathscr{H} H
ω ˙ = { ω , H } \dot{\omega}=\{\omega,\mathscr{H}\} ω ˙ = { ω , H } 因此，对于变换后的速度，有
q ˉ ˙ j = { q ˉ j , H } = ∑ i ( ∂ q ˉ j ∂ q i ∂ H ∂ p i − ∂ q ˉ j ∂ p i ∂ H ∂ q i ) = ∑ i ∑ k [ ∂ q ˉ j ∂ q i ( ∂ H ∂ q ˉ k ∂ q ˉ k ∂ p i + ∂ H ∂ p ˉ k ∂ p ˉ k ∂ p i ) − ∂ q ˉ j ∂ p i ( ∂ H ∂ q ˉ k ∂ q ˉ k ∂ q i + ∂ H ∂ p ˉ k ∂ p ˉ k ∂ q i ) ] = ∑ k ∂ H ∂ q ˉ k ∑ i ( ∂ q ˉ j ∂ q i ∂ q ˉ k ∂ p i − ∂ q ˉ j ∂ p i ∂ q ˉ k ∂ q i ) + ∑ k ∂ H ∂ p ˉ k ∑ i ( ∂ q ˉ j ∂ q i ∂ p ˉ k ∂ p i − ∂ q ˉ j ∂ p i ∂ p ˉ k ∂ q i ) = ∑ k ∂ H ∂ q ˉ k { q ˉ j , q ˉ k } + ∑ k ∂ H ∂ p ˉ k { q ˉ j , p ˉ k } \begin{aligned}
\dot{\bar{q}}_{j}&=\{\bar{q}_{j},\mathscr{H}\}\\
&=\sum_{i}\left(\frac{\partial \bar{q}_j}{\partial q_i} \frac{\partial \mathscr{H}}{\partial p_i}-\frac{\partial \bar{q}_j}{\partial p_i} \frac{\partial \mathscr{H}}{\partial q_i}\right)\\
&=\sum_{i}\sum_{k}\left[\frac{\partial \bar{q}_j}{\partial q_i}\left(\frac{\partial \mathscr{H}}{\partial \bar{q}_{k}}\frac{\partial \bar{q}_{k}}{\partial p_{i}}+\frac{\partial \mathscr{H}}{\partial \bar{p}_{k}}\frac{\partial \bar{p}_{k}}{\partial p_{i}}\right)-\frac{\partial \bar{q}_j}{\partial p_i}\left(\frac{\partial \mathscr{H}}{\partial \bar{q}_{k}}\frac{\partial \bar{q}_{k}}{\partial q_{i}}+\frac{\partial \mathscr{H}}{\partial \bar{p}_{k}}\frac{\partial \bar{p}_{k}}{\partial q_{i}}\right)\right]\\
&=\sum_{k}\frac{\partial \mathscr{H}}{\partial \bar{q}_{k}}\sum_{i}\left(\frac{\partial \bar{q}_j}{\partial q_i}\frac{\partial \bar{q}_{k}}{\partial p_{i}}-\frac{\partial \bar{q}_j}{\partial p_i}\frac{\partial \bar{q}_{k}}{\partial q_{i}}\right)+\sum_{k}\frac{\partial \mathscr{H}}{\partial \bar{p}_{k}}\sum_{i}\left(\frac{\partial \bar{q}_j}{\partial q_i}\frac{\partial \bar{p}_{k}}{\partial p_{i}}-\frac{\partial \bar{q}_j}{\partial p_i}\frac{\partial \bar{p}_{k}}{\partial q_{i}}\right)\\
&=\sum_{k}\frac{\partial \mathscr{H}}{\partial \bar{q}_{k}}\{\bar{q}_{j},\bar{q}_{k}\}+\sum_{k}\frac{\partial \mathscr{H}}{\partial \bar{p}_{k}}\{\bar{q}_{j},\bar{p}_{k}\}
\end{aligned} q ˉ ˙ j = { q ˉ j , H } = i ∑ ( ∂ q i ∂ q ˉ j ∂ p i ∂ H − ∂ p i ∂ q ˉ j ∂ q i ∂ H ) = i ∑ k ∑ [ ∂ q i ∂ q ˉ j ( ∂ q ˉ k ∂ H ∂ p i ∂ q ˉ k + ∂ p ˉ k ∂ H ∂ p i ∂ p ˉ k ) − ∂ p i ∂ q ˉ j ( ∂ q ˉ k ∂ H ∂ q i ∂ q ˉ k + ∂ p ˉ k ∂ H ∂ q i ∂ p ˉ k ) ] = k ∑ ∂ q ˉ k ∂ H i ∑ ( ∂ q i ∂ q ˉ j ∂ p i ∂ q ˉ k − ∂ p i ∂ q ˉ j ∂ q i ∂ q ˉ k ) + k ∑ ∂ p ˉ k ∂ H i ∑ ( ∂ q i ∂ q ˉ j ∂ p i ∂ p ˉ k − ∂ p i ∂ q ˉ j ∂ q i ∂ p ˉ k ) = k ∑ ∂ q ˉ k ∂ H { q ˉ j , q ˉ k } + k ∑ ∂ p ˉ k ∂ H { q ˉ j , p ˉ k } 为了满足Hamilton方程，必须有
{ q ˉ j , q ˉ k } = 0 { q ˉ j , p ˉ k } = δ j k \begin{aligned}
\{\bar{q}_{j},\bar{q}_{k}\}&=0\\
\{\bar{q}_{j},\bar{p}_{k}\}&=\delta_{jk}
\end{aligned} { q ˉ j , q ˉ k } { q ˉ j , p ˉ k } = 0 = δ j k 对变换后动量的时间导数也可以作同样的计算
p ˉ ˙ j = { p ˉ j , H } = ∑ i ( ∂ p ˉ j ∂ q i ∂ H ∂ p i − ∂ p ˉ j ∂ p i ∂ H ∂ q i ) = ∑ i ∑ k [ ∂ p ˉ j ∂ q i ( ∂ H ∂ q ˉ k ∂ q ˉ k ∂ p i + ∂ H ∂ p ˉ k ∂ p ˉ k ∂ p i ) − ∂ p ˉ j ∂ p i ( ∂ H ∂ q ˉ k ∂ q ˉ k ∂ q i + ∂ H ∂ p ˉ k ∂ p ˉ k ∂ q i ) ] = ∑ k ∂ H ∂ q ˉ k ∑ i ( ∂ p ˉ j ∂ q i ∂ q ˉ k ∂ p i − ∂ p ˉ j ∂ p i ∂ q ˉ k ∂ q i ) + ∑ k ∂ H ∂ p ˉ k ∑ i ( ∂ p ˉ j ∂ q i ∂ p ˉ k ∂ p i − ∂ p ˉ j ∂ p i ∂ p ˉ k ∂ q i ) = ∑ k ∂ H ∂ q ˉ k { p ˉ j , q ˉ k } + ∑ k ∂ H ∂ p ˉ k { p ˉ j , p ˉ k } \begin{aligned}
\dot{\bar{p}}_{j}&=\{\bar{p}_{j},\mathscr{H}\}\\
&=\sum_{i}\left(\frac{\partial \bar{p}_{j}}{\partial q_i} \frac{\partial \mathscr{H}}{\partial p_i}-\frac{\partial \bar{p}_{j}}{\partial p_i} \frac{\partial \mathscr{H}}{\partial q_i}\right)\\
&=\sum_{i}\sum_{k}\left[\frac{\partial \bar{p}_{j}}{\partial q_i}\left(\frac{\partial \mathscr{H}}{\partial \bar{q}_{k}}\frac{\partial \bar{q}_{k}}{\partial p_{i}}+\frac{\partial \mathscr{H}}{\partial \bar{p}_{k}}\frac{\partial \bar{p}_{k}}{\partial p_{i}}\right)-\frac{\partial \bar{p}_{j}}{\partial p_i}\left(\frac{\partial \mathscr{H}}{\partial \bar{q}_{k}}\frac{\partial \bar{q}_{k}}{\partial q_{i}}+\frac{\partial \mathscr{H}}{\partial \bar{p}_{k}}\frac{\partial \bar{p}_{k}}{\partial q_{i}}\right)\right]\\
&=\sum_{k}\frac{\partial \mathscr{H}}{\partial \bar{q}_{k}}\sum_{i}\left(\frac{\partial \bar{p}_{j}}{\partial q_i}\frac{\partial \bar{q}_{k}}{\partial p_{i}}-\frac{\partial \bar{p}_{j}}{\partial p_i}\frac{\partial \bar{q}_{k}}{\partial q_{i}}\right)+\sum_{k}\frac{\partial \mathscr{H}}{\partial \bar{p}_{k}}\sum_{i}\left(\frac{\partial \bar{p}_{j}}{\partial q_i}\frac{\partial \bar{p}_{k}}{\partial p_{i}}-\frac{\partial \bar{p}_{j}}{\partial p_i}\frac{\partial \bar{p}_{k}}{\partial q_{i}}\right)\\
&=\sum_{k}\frac{\partial \mathscr{H}}{\partial \bar{q}_{k}}\{\bar{p}_{j},\bar{q}_{k}\}+\sum_{k}\frac{\partial \mathscr{H}}{\partial \bar{p}_{k}}\{\bar{p}_{j},\bar{p}_{k}\}
\end{aligned} p ˉ ˙ j = { p ˉ j , H } = i ∑ ( ∂ q i ∂ p ˉ j ∂ p i ∂ H − ∂ p i ∂ p ˉ j ∂ q i ∂ H ) = i ∑ k ∑ [ ∂ q i ∂ p ˉ j ( ∂ q ˉ k ∂ H ∂ p i ∂ q ˉ k + ∂ p ˉ k ∂ H ∂ p i ∂ p ˉ k ) − ∂ p i ∂ p ˉ j ( ∂ q ˉ k ∂ H ∂ q i ∂ q ˉ k + ∂ p ˉ k ∂ H ∂ q i ∂ p ˉ k ) ] = k ∑ ∂ q ˉ k ∂ H i ∑ ( ∂ q i ∂ p ˉ j ∂ p i ∂ q ˉ k − ∂ p i ∂ p ˉ j ∂ q i ∂ q ˉ k ) + k ∑ ∂ p ˉ k ∂ H i ∑ ( ∂ q i ∂ p ˉ j ∂ p i ∂ p ˉ k − ∂ p i ∂ p ˉ j ∂ q i ∂ p ˉ k ) = k ∑ ∂ q ˉ k ∂ H { p ˉ j , q ˉ k } + k ∑ ∂ p ˉ k ∂ H { p ˉ j , p ˉ k } 为了满足Hamilton方程，必须有
{ p ˉ j , q ˉ k } = − δ j k { p ˉ j , p ˉ k } = 0 \begin{aligned}
\{\bar{p}_{j},\bar{q}_{k}\}&=-\delta_{jk}\\
\{\bar{p}_{j},\bar{p}_{k}\}&=0
\end{aligned} { p ˉ j , q ˉ k } { p ˉ j , p ˉ k } = − δ j k = 0 因此可以得出，为使变换成为正则变换，条件为
{ q ˉ j , q ˉ k } = { p ˉ j , p ˉ k } = 0 { q ˉ j , p ˉ k } = δ j k \begin{aligned}
\{\bar{q}_{j},\bar{q}_{k}\}&=\{\bar{p}_{j},\bar{p}_{k}\}=0\\
\{\bar{q}_{j},\bar{p}_{k}\}&=\delta_{jk}
\end{aligned} { q ˉ j , q ˉ k } { q ˉ j , p ˉ k } = { p ˉ j , p ˉ k } = 0 = δ j k ■ ~\tag*{$\blacksquare$} ■ 习题 2.7.4
验证到旋转坐标系的变换
x ˉ = x cos θ − y sin θ y ˉ = x sin θ + y cos θ p ˉ x = p x cos θ − p y sin θ p ˉ y = p x sin θ + p y cos θ \begin{gathered}
\bar{x}=x \cos \theta-y \sin \theta \\
\bar{y}=x \sin \theta+y \cos \theta \\
\bar{p}_x=p_x \cos \theta-p_y \sin \theta \\
\bar{p}_y=p_x \sin \theta+p_y \cos \theta
\end{gathered} x ˉ = x cos θ − y sin θ y ˉ = x sin θ + y cos θ p ˉ x = p x cos θ − p y sin θ p ˉ y = p x sin θ + p y cos θ 是一个正则变换。
解答:
为了证明这是正则变换，我们必须计算Poisson括号。在计算Poisson括号之前，可以先求出所有非零偏导数
∂ x ˉ ∂ x = cos θ ∂ x ˉ ∂ y = − sin θ ∂ y ˉ ∂ x = sin θ ∂ y ˉ ∂ y = cos θ ∂ p ˉ x ∂ p x = cos θ ∂ p ˉ x ∂ p y = − sin θ ∂ p ˉ y ∂ p x = sin θ ∂ p ˉ y ∂ p y = cos θ \begin{aligned}
\frac{\partial\bar{x}}{\partial x}=\cos\theta\qquad &\frac{\partial\bar{x}}{\partial y}=-\sin\theta\\
\frac{\partial\bar{y}}{\partial x}=\sin\theta\qquad &\frac{\partial\bar{y}}{\partial y}=\cos\theta\\
\frac{\partial\bar{p}_{x}}{\partial p_{x}}=\cos\theta\qquad &\frac{\partial\bar{p}_{x}}{\partial p_{y}}=-\sin\theta\\
\frac{\partial\bar{p}_{y}}{\partial p_{x}}=\sin\theta\qquad &\frac{\partial\bar{p}_{y}}{\partial p_{y}}=\cos\theta\\
\end{aligned} ∂ x ∂ x ˉ = cos θ ∂ x ∂ y ˉ = sin θ ∂ p x ∂ p ˉ x = cos θ ∂ p x ∂ p ˉ y = sin θ ∂ y ∂ x ˉ = − sin θ ∂ y ∂ y ˉ = cos θ ∂ p y ∂ p ˉ x = − sin θ ∂ p y ∂ p ˉ y = cos θ 其中 q 1 = x q_{1}=x q 1 = x q 2 = y q_{2}=y q 2 = y p 1 = p x p_{1}=p_{x} p 1 = p x p 2 = p y p_{2}=p_{y} p 2 = p y
{ x ˉ , y ˉ } = ∑ i ( ∂ x ˉ ∂ q i ∂ y ˉ ∂ p i − ∂ x ˉ ∂ p i ∂ y ˉ ∂ q i ) = 0 \{\bar{x},\bar{y}\}=\sum_{i}\left(\frac{\partial \bar{x}}{\partial q_i} \frac{\partial \bar{y}}{\partial p_i}-\frac{\partial \bar{x}}{\partial p_i} \frac{\partial \bar{y}}{\partial q_i}\right)=0 { x ˉ , y ˉ } = i ∑ ( ∂ q i ∂ x ˉ ∂ p i ∂ y ˉ − ∂ p i ∂ x ˉ ∂ q i ∂ y ˉ ) = 0 因为两个坐标都不依赖于任何动量。类似地，
{ p ˉ x , p ˉ y } = 0 \{\bar{p}_{x},\bar{p}_{y}\}=0 { p ˉ x , p ˉ y } = 0 因为Poisson括号中包含 p ˉ i \bar{p}_{i} p ˉ i q i q_{i} q i
其余Poisson括号均形如 { q ˉ i , p ˉ j } \{\bar{q}_{i},\bar{p}_{j}\} { q ˉ i , p ˉ j }
{ x ˉ , p ˉ x } = ∑ i ( ∂ x ˉ ∂ q i ∂ p ˉ x ∂ p i − ∂ x ˉ ∂ p i ∂ p ˉ x ∂ q i ) = ∂ x ˉ ∂ x ∂ p ˉ x ∂ p x + ∂ x ˉ ∂ y ∂ p ˉ x ∂ p y = cos 2 θ + sin 2 θ = 1 { x ˉ , p ˉ y } = ∑ i ( ∂ x ˉ ∂ q i ∂ p ˉ y ∂ p i − ∂ x ˉ ∂ p i ∂ p ˉ y ∂ q i ) = ∂ x ˉ ∂ x ∂ p ˉ y ∂ p x + ∂ x ˉ ∂ y ∂ p ˉ y ∂ p y = sin θ cos θ − sin θ cos θ = 0 \begin{aligned}
\{\bar{x},\bar{p}_{x}\}& =\sum_i\left(\frac{\partial \bar{x}}{\partial q_i} \frac{\partial \bar{p}_x}{\partial p_i}-\frac{\partial \bar{x}}{\partial p_i} \frac{\partial \bar{p}_x}{\partial q_i}\right) \\
& =\frac{\partial \bar{x}}{\partial x} \frac{\partial \bar{p}_x}{\partial p_x}+\frac{\partial \bar{x}}{\partial y} \frac{\partial \bar{p}_x}{\partial p_y} \\
& =\cos ^2 \theta+\sin ^2 \theta \\
& =1\\
\left\{\bar{x}, \bar{p}_y\right\} & =\sum_i\left(\frac{\partial \bar{x}}{\partial q_i} \frac{\partial \bar{p}_y}{\partial p_i}-\frac{\partial \bar{x}}{\partial p_i} \frac{\partial \bar{p}_y}{\partial q_i}\right) \\
& =\frac{\partial \bar{x}}{\partial x} \frac{\partial \bar{p}_y}{\partial p_x}+\frac{\partial \bar{x}}{\partial y} \frac{\partial \bar{p}_y}{\partial p_y} \\
& =\sin \theta \cos \theta-\sin \theta \cos \theta \\
& =0
\end{aligned} { x ˉ , p ˉ x } { x ˉ , p ˉ y } = i ∑ ( ∂ q i ∂ x ˉ ∂ p i ∂ p ˉ x − ∂ p i ∂ x ˉ ∂ q i ∂ p ˉ x ) = ∂ x ∂ x ˉ ∂ p x ∂ p ˉ x + ∂ y ∂ x ˉ ∂ p y ∂ p ˉ x = cos 2 θ + sin 2 θ = 1 = i ∑ ( ∂ q i ∂ x ˉ ∂ p i ∂ p ˉ y − ∂ p i ∂ x ˉ ∂ q i ∂ p ˉ y ) = ∂ x ∂ x ˉ ∂ p x ∂ p ˉ y + ∂ y ∂ x ˉ ∂ p y ∂ p ˉ y = sin θ cos θ − sin θ cos θ = 0 类似地，
{ y ˉ , p ˉ x } = ∑ i ( ∂ y ˉ ∂ q i ∂ p ˉ x ∂ p i − ∂ y ˉ ∂ p i ∂ p ˉ x ∂ q i ) = ∂ y ˉ ∂ x ∂ p ˉ x ∂ p x + ∂ y ˉ ∂ y ∂ p ˉ x ∂ p y = sin θ cos θ + cos θ ( − sin θ ) = 0 { y ˉ , p ˉ y } = ∑ i ( ∂ y ˉ ∂ q i ∂ p ˉ y ∂ p i − ∂ y ˉ ∂ p i ∂ p ˉ y ∂ q i ) = ∂ y ˉ ∂ x ∂ p ˉ y ∂ p x + ∂ y ˉ ∂ y ∂ p ˉ y ∂ p y = sin θ sin θ + cos θ cos θ = 1 \begin{aligned}
\left\{\bar{y}, \bar{p}_x\right\}&=\sum_i\left(\frac{\partial \bar{y}}{\partial q_i} \frac{\partial \bar{p}_x}{\partial p_i}-\frac{\partial \bar{y}}{\partial p_i} \frac{\partial \bar{p}_x}{\partial q_i}\right) \\
&=\frac{\partial \bar{y}}{\partial x} \frac{\partial \bar{p}_x}{\partial p_x}+\frac{\partial \bar{y}}{\partial y} \frac{\partial \bar{p}_x}{\partial p_y}\\
&=\sin\theta\cos\theta+\cos\theta(-\sin\theta)\\
&=0 \\
\left\{\bar{y}, \bar{p}_y\right\}&=\sum_i\left(\frac{\partial \bar{y}}{\partial q_i} \frac{\partial \bar{p}_y}{\partial p_i}-\frac{\partial \bar{y}}{\partial p_i} \frac{\partial \bar{p}_y}{\partial q_i}\right) \\
&=\frac{\partial \bar{y}}{\partial x} \frac{\partial \bar{p}_y}{\partial p_x}+\frac{\partial \bar{y}}{\partial y} \frac{\partial \bar{p}_y}{\partial p_y}\\
&=\sin\theta\sin\theta+\cos\theta\cos\theta\\
&=1
\end{aligned} { y ˉ , p ˉ x } { y ˉ , p ˉ y } = i ∑ ( ∂ q i ∂ y ˉ ∂ p i ∂ p ˉ x − ∂ p i ∂ y ˉ ∂ q i ∂ p ˉ x ) = ∂ x ∂ y ˉ ∂ p x ∂ p ˉ x + ∂ y ∂ y ˉ ∂ p y ∂ p ˉ x = sin θ cos θ + cos θ ( − sin θ ) = 0 = i ∑ ( ∂ q i ∂ y ˉ ∂ p i ∂ p ˉ y − ∂ p i ∂ y ˉ ∂ q i ∂ p ˉ y ) = ∂ x ∂ y ˉ ∂ p x ∂ p ˉ y + ∂ y ∂ y ˉ ∂ p y ∂ p ˉ y = sin θ sin θ + cos θ cos θ = 1 因此，旋转坐标系的变换是一个正则变换。
■ ~\tag*{$\blacksquare$} ■ 习题 2.7.5
证明极坐标变量
ρ = ( x 2 + y 2 ) 1 / 2 , ϕ = tan − 1 ( y / x ) p ρ = e ^ ρ ⋅ p = x p x + y p y ( x 2 + y 2 ) 1 / 2 , p ϕ = x p y − y p x ( = l z ) \begin{aligned}
\rho=\left(x^2+y^2\right)^{1 / 2},\quad &\phi=\tan^{-1}(y / x)\\
p_\rho=\hat{e}_\rho \cdot \mathbf{p}=\frac{x p_x+y p_y}{\left(x^2+y^2\right)^{1 / 2}}, \quad & p_\phi=x p_y-y p_x\left(=l_z\right)
\end{aligned} ρ = ( x 2 + y 2 ) 1/2 , p ρ = e ^ ρ ⋅ p = ( x 2 + y 2 ) 1/2 x p x + y p y , ϕ = tan − 1 ( y / x ) p ϕ = x p y − y p x ( = l z ) 是正则变量。（e ^ ρ \hat{e}_\rho e ^ ρ
解答:
非零偏导数为
∂ ρ ∂ x = x x 2 + y 2 ∂ ρ ∂ y = y x 2 + y 2 ∂ ϕ ∂ x = − y x 2 + y 2 ∂ ϕ ∂ y = x x 2 + y 2 ∂ p ρ ∂ x = y 2 p x − x y p y ( x 2 + y 2 ) 3 / 2 ∂ p ρ ∂ y = x 2 p y − x y p x ( x 2 + y 2 ) 3 / 2 ∂ p ρ ∂ p x = x x 2 + y 2 ∂ p ρ ∂ p y = y x 2 + y 2 ∂ p ϕ ∂ x = p y ∂ p ϕ ∂ y = − p x ∂ p ϕ ∂ p x = − y ∂ p ϕ ∂ p y = x \begin{aligned}
&\frac{\partial\rho}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}}} &\quad &\frac{\partial \rho}{\partial y}=\frac{y}{\sqrt{x^{2}+y^{2}}}\\
&\frac{\partial\phi}{\partial x}=\frac{-y}{x^{2}+y^{2}} &\quad &\frac{\partial \phi}{\partial y}=\frac{x}{x^{2}+y^{2}}\\
&\frac{\partial p_{\rho}}{\partial x}=\frac{y^{2}p_{x}-xyp_{y}}{(x^{2}+y^{2})^{3/2}} &\quad &\frac{\partial p_{\rho}}{\partial y}=\frac{x^{2}p_{y}-xyp_{x}}{(x^{2}+y^{2})^{3/2}}\\
&\frac{\partial p_{\rho}}{\partial p_{x}}=\frac{x}{\sqrt{x^{2}+y^{2}}} &\quad & \frac{\partial p_{\rho}}{\partial p_{y}}=\frac{y}{\sqrt{x^{2}+y^{2}}}\\
&\frac{\partial p_{\phi}}{\partial x}=p_{y} &\quad & \frac{\partial p_{\phi}}{\partial y}=-p_{x}\\
&\frac{\partial p_{\phi}}{\partial p_{x}}=-y &\quad & \frac{\partial p_{\phi}}{\partial p_{y}}=x
\end{aligned} ∂ x ∂ ρ = x 2 + y 2 x ∂ x ∂ ϕ = x 2 + y 2 − y ∂ x ∂ p ρ = ( x 2 + y 2 ) 3/2 y 2 p x − x y p y ∂ p x ∂ p ρ = x 2 + y 2 x ∂ x ∂ p ϕ = p y ∂ p x ∂ p ϕ = − y ∂ y ∂ ρ = x 2 + y 2 y ∂ y ∂ ϕ = x 2 + y 2 x ∂ y ∂ p ρ = ( x 2 + y 2 ) 3/2 x 2 p y − x y p x ∂ p y ∂ p ρ = x 2 + y 2 y ∂ y ∂ p ϕ = − p x ∂ p y ∂ p ϕ = x 现在计算Poisson括号
{ ρ , ϕ } = ∑ i ( ∂ ρ ∂ q i ∂ ϕ ∂ p i − ∂ ρ ∂ p i ∂ ϕ ∂ q i ) = 0 \{\rho,\phi\}=\sum_i\left(\frac{\partial \rho}{\partial q_i} \frac{\partial \phi}{\partial p_i}-\frac{\partial \rho}{\partial p_i} \frac{\partial \phi}{\partial q_i}\right)=0 { ρ , ϕ } = i ∑ ( ∂ q i ∂ ρ ∂ p i ∂ ϕ − ∂ p i ∂ ρ ∂ q i ∂ ϕ ) = 0 因为坐标不依赖于动量。
{ p ρ , p ϕ } = ∑ i ( ∂ p ρ ∂ q i ∂ p ϕ ∂ p i − ∂ p ρ ∂ p i ∂ p ϕ ∂ q i ) = ∂ p ρ ∂ x ∂ p ϕ ∂ p x − ∂ p ρ ∂ p x ∂ p ϕ ∂ x + ∂ p ρ ∂ y ∂ p ϕ ∂ p y − ∂ p ρ ∂ p y ∂ p ϕ ∂ y = y 2 p x − x y p y ( x 2 + y 2 ) 3 / 2 ( − y ) − x x 2 + y 2 p y + x 2 p y − x y p x ( x 2 + y 2 ) 3 / 2 x − y x 2 + y 2 ( − p x ) = − y 3 p x + x y 2 p y − ( x 3 + x y 2 ) p y + x 3 p y − x 2 y p x + ( x 2 y + y 3 ) p x ( x 2 + y 2 ) 3 / 2 = 0 \begin{aligned}
\{p_{\rho},p_{\phi}\}&=\sum_{i}\left(\frac{\partial p_\rho}{\partial q_i} \frac{\partial p_\phi}{\partial p_i}-\frac{\partial p_\rho}{\partial p_i} \frac{\partial p_\phi}{\partial q_i}\right)\\
&=\frac{\partial p_\rho}{\partial x} \frac{\partial p_\phi}{\partial p_x}-\frac{\partial p_\rho}{\partial p_x} \frac{\partial p_\phi}{\partial x}+\frac{\partial p_\rho}{\partial y} \frac{\partial p_\phi}{\partial p_y}-\frac{\partial p_\rho}{\partial p_y} \frac{\partial p_\phi}{\partial y}\\
&=\frac{y^{2}p_{x}-xyp_{y}}{(x^{2}+y^{2})^{3/2}}(-y)-\frac{x}{\sqrt{x^{2}+y^{2}}}p_{y}+\frac{x^{2}p_{y}-xyp_{x}}{(x^{2}+y^{2})^{3/2}}x-\frac{y}{\sqrt{x^{2}+y^{2}}}(-p_{x})\\
&=\frac{-y^{3}p_{x}+xy^{2}p_{y}-(x^{3}+xy^{2})p_{y}+x^{3}p_{y}-x^{2}yp_{x}+(x^{2}y+y^{3})p_{x}}{(x^{2}+y^{2})^{3/2}}\\
&=0
\end{aligned} { p ρ , p ϕ } = i ∑ ( ∂ q i ∂ p ρ ∂ p i ∂ p ϕ − ∂ p i ∂ p ρ ∂ q i ∂ p ϕ ) = ∂ x ∂ p ρ ∂ p x ∂ p ϕ − ∂ p x ∂ p ρ ∂ x ∂ p ϕ + ∂ y ∂ p ρ ∂ p y ∂ p ϕ − ∂ p y ∂ p ρ ∂ y ∂ p ϕ = ( x 2 + y 2 ) 3/2 y 2 p x − x y p y ( − y ) − x 2 + y 2 x p y + ( x 2 + y 2 ) 3/2 x 2 p y − x y p x x − x 2 + y 2 y ( − p x ) = ( x 2 + y 2 ) 3/2 − y 3 p x + x y 2 p y − ( x 3 + x y 2 ) p y + x 3 p y − x 2 y p x + ( x 2 y + y 3 ) p x = 0 其余Poisson括号均形如 { q ˉ i , p ˉ j } \{\bar{q}_{i},\bar{p}_{j}\} { q ˉ i , p ˉ j }
{ ρ , p ρ } = ∑ i ( ∂ ρ ∂ q i ∂ p ρ ∂ p i − ∂ ρ ∂ p i ∂ p ρ ∂ q i ) = ∂ ρ ∂ x ∂ p ρ ∂ p x − ∂ ρ ∂ p x ∂ p ρ ∂ x + ∂ ρ ∂ y ∂ p ρ ∂ p y − ∂ ρ ∂ p y ∂ p ρ ∂ y = x 2 x 2 + y 2 − 0 + y 2 x 2 + y 2 − 0 = 1 { ρ , p ϕ } = ∑ i ( ∂ ρ ∂ q i ∂ p ϕ ∂ p i − ∂ ρ ∂ p i ∂ p ϕ ∂ q i ) = ∂ ρ ∂ x ∂ p ϕ ∂ p x − ∂ ρ ∂ p x ∂ p ϕ ∂ x + ∂ ρ ∂ y ∂ p ϕ ∂ p y − ∂ ρ ∂ p y ∂ p ϕ ∂ y = − x y x 2 + y 2 − 0 + x y x 2 + y 2 − 0 = 0 { ϕ , p ρ } = ∑ i ( ∂ ϕ ∂ q i ∂ p ρ ∂ p i − ∂ ϕ ∂ p i ∂ p ρ ∂ q i ) = ∂ ϕ ∂ x ∂ p ρ ∂ p x − ∂ ϕ ∂ p x ∂ p ρ ∂ x + ∂ ϕ ∂ y ∂ p ρ ∂ p y − ∂ ϕ ∂ p y ∂ p ρ ∂ y = − y x 2 + y 2 x x 2 + y 2 − 0 + x x 2 + y 2 y x 2 + y 2 − 0 = 0 { ϕ , p ϕ } = ∑ i ( ∂ ϕ ∂ q i ∂ p ϕ ∂ p i − ∂ ϕ ∂ p i ∂ p ϕ ∂ q i ) = ∂ ϕ ∂ x ∂ p ϕ ∂ p x − ∂ ϕ ∂ p x ∂ p ϕ ∂ x + ∂ ϕ ∂ y ∂ p ϕ ∂ p y − ∂ ϕ ∂ p y ∂ p ϕ ∂ y = − y x 2 + y 2 ( − y ) − 0 + x x 2 + y 2 x − 0 = 1 \begin{aligned}
\left\{\rho, p_\rho\right\}&=\sum_i\left(\frac{\partial \rho}{\partial q_i} \frac{\partial p_\rho}{\partial p_i}-\frac{\partial \rho}{\partial p_i} \frac{\partial p_\rho}{\partial q_i}\right) \\
&=\frac{\partial \rho}{\partial x} \frac{\partial p_\rho}{\partial p_x}-\frac{\partial \rho}{\partial p_x} \frac{\partial p_\rho}{\partial x}+\frac{\partial \rho}{\partial y} \frac{\partial p_\rho}{\partial p_y}-\frac{\partial \rho}{\partial p_y} \frac{\partial p_\rho}{\partial y}\\
& =\frac{x^2}{x^2+y^2}-0+\frac{y^2}{x^2+y^2}-0 \\
& =1 \\
\left\{\rho, p_\phi\right\}&=\sum_i\left(\frac{\partial \rho}{\partial q_i} \frac{\partial p_\phi}{\partial p_i}-\frac{\partial \rho}{\partial p_i} \frac{\partial p_\phi}{\partial q_i}\right) \\
&=\frac{\partial \rho}{\partial x} \frac{\partial p_\phi}{\partial p_x}-\frac{\partial \rho}{\partial p_x} \frac{\partial p_\phi}{\partial x}+\frac{\partial \rho}{\partial y} \frac{\partial p_\phi}{\partial p_y}-\frac{\partial \rho}{\partial p_y} \frac{\partial p_\phi}{\partial y}\\
& =-\frac{x y}{\sqrt{x^2+y^2}}-0+\frac{x y}{\sqrt{x^2+y^2}}-0 \\
& =0 \\
\left\{\phi, p_\rho\right\}&=\sum_i\left(\frac{\partial \phi}{\partial q_i} \frac{\partial p_\rho}{\partial p_i}-\frac{\partial \phi}{\partial p_i} \frac{\partial p_\rho}{\partial q_i}\right) \\
&=\frac{\partial \phi}{\partial x} \frac{\partial p_\rho}{\partial p_x}-\frac{\partial \phi}{\partial p_x} \frac{\partial p_\rho}{\partial x}+\frac{\partial \phi}{\partial y} \frac{\partial p_\rho}{\partial p_y}-\frac{\partial \phi}{\partial p_y} \frac{\partial p_\rho}{\partial y}\\
& =\frac{-y}{x^2+y^2}\frac{x}{\sqrt{x^{2}+y^{2}}}-0+\frac{x}{x^2+y^2}\frac{y}{\sqrt{x^{2}+y^{2}}}-0 \\
& =0 \\
\left\{\phi, p_\phi\right\}&=\sum_i\left(\frac{\partial \phi}{\partial q_i} \frac{\partial p_\phi}{\partial p_i}-\frac{\partial \phi}{\partial p_i} \frac{\partial p_\phi}{\partial q_i}\right) \\
&=\frac{\partial \phi}{\partial x} \frac{\partial p_\phi}{\partial p_x}-\frac{\partial \phi}{\partial p_x} \frac{\partial p_\phi}{\partial x}+\frac{\partial \phi}{\partial y} \frac{\partial p_\phi}{\partial p_y}-\frac{\partial \phi}{\partial p_y} \frac{\partial p_\phi}{\partial y}\\
&=\frac{-y}{x^2+y^2}(-y)-0+\frac{x}{x^2+y^2}x-0\\
&=1
\end{aligned} { ρ , p ρ } { ρ , p ϕ } { ϕ , p ρ } { ϕ , p ϕ } = i ∑ ( ∂ q i ∂ ρ ∂ p i ∂ p ρ − ∂ p i ∂ ρ ∂ q i ∂ p ρ ) = ∂ x ∂ ρ ∂ p x ∂ p ρ − ∂ p x ∂ ρ ∂ x ∂ p ρ + ∂ y ∂ ρ ∂ p y ∂ p ρ − ∂ p y ∂ ρ ∂ y ∂ p ρ = x 2 + y 2 x 2 − 0 + x 2 + y 2 y 2 − 0 = 1 = i ∑ ( ∂ q i ∂ ρ ∂ p i ∂ p ϕ − ∂ p i ∂ ρ ∂ q i ∂ p ϕ ) = ∂ x ∂ ρ ∂ p x ∂ p ϕ − ∂ p x ∂ ρ ∂ x ∂ p ϕ + ∂ y ∂ ρ ∂ p y ∂ p ϕ − ∂ p y ∂ ρ ∂ y ∂ p ϕ = − x 2 + y 2 x y − 0 + x 2 + y 2 x y − 0 = 0 = i ∑ ( ∂ q i ∂ ϕ ∂ p i ∂ p ρ − ∂ p i ∂ ϕ ∂ q i ∂ p ρ ) = ∂ x ∂ ϕ ∂ p x ∂ p ρ − ∂ p x ∂ ϕ ∂ x ∂ p ρ + ∂ y ∂ ϕ ∂ p y ∂ p ρ − ∂ p y ∂ ϕ ∂ y ∂ p ρ = x 2 + y 2 − y x 2 + y 2 x − 0 + x 2 + y 2 x x 2 + y 2 y − 0 = 0 = i ∑ ( ∂ q i ∂ ϕ ∂ p i ∂ p ϕ − ∂ p i ∂ ϕ ∂ q i ∂ p ϕ ) = ∂ x ∂ ϕ ∂ p x ∂ p ϕ − ∂ p x ∂ ϕ ∂ x ∂ p ϕ + ∂ y ∂ ϕ ∂ p y ∂ p ϕ − ∂ p y ∂ ϕ ∂ y ∂ p ϕ = x 2 + y 2 − y ( − y ) − 0 + x 2 + y 2 x x − 0 = 1 因此所有Poisson括号都满足要求，所以该变换是正则变换。
■ ~\tag*{$\blacksquare$} ■ 习题 2.7.6
验证从变量 r 1 , r 2 , p 1 , p 2 \mathbf{r}_1, \mathbf{r}_2, \mathbf{p}_1, \mathbf{p}_2 r 1 , r 2 , p 1 , p 2 r C M , p C M , r \mathbf{r}_{\mathrm{CM}}, \mathbf{p}_{\mathrm{CM}}, \mathbf{r} r CM , p CM , r p \mathbf{p} p
解答:
从质量分别为 m 1 m_{1} m 1 m 2 m_{2} m 2 r 1 \mathbf{r}_{1} r 1 r 2 \mathbf{r}_{2} r 2 r \mathbf{r} r r C M \mathbf{r}_{CM} r C M
r = r 1 − r 2 r C M = m 1 r 1 + m 2 r 2 M \begin{aligned}
\mathbf{r}&=\mathbf{r}_{1}-\mathbf{r}_{2}\\
\mathbf{r}_{CM}&=\frac{m_{1}\mathbf{r}_{1}+m_{2}\mathbf{r}_{2}}{M}
\end{aligned} r r C M = r 1 − r 2 = M m 1 r 1 + m 2 r 2 其中 M : = m 1 + m 2 M:=m_{1}+m_{2} M := m 1 + m 2
p i = m i r ˙ i \mathbf{p}_{i}=m_{i}\dot{\mathbf{r}}_{i} p i = m i r ˙ i 共轭动量按如下方式变换
p = μ r ˙ = m 2 p 1 − m 1 p 2 M p C M = M r ˙ C M = p 1 + p 2 \begin{aligned}
\mathbf{p}&=\mu\dot{\mathbf{r}}=\frac{m_{2}\mathbf{p}_{1}-m_{1}\mathbf{p}_{2}}{M}\\
\mathbf{p}_{CM}&=M\dot{\mathbf{r}}_{CM}=\mathbf{p}_{1}+\mathbf{p}_{2}
\end{aligned} p p C M = μ r ˙ = M m 2 p 1 − m 1 p 2 = M r ˙ C M = p 1 + p 2 其中 μ : = m 1 m 2 M \mu:=\frac{m_{1}m_{2}}{M} μ := M m 1 m 2
现在计算Poisson括号，以检验该变换是否为正则变换。
注意，新坐标只依赖于旧坐标；反过来，新动量只依赖于旧动量。还要注意，r i r_{i} r i r 1 \mathbf{r}_{1} r 1 r 2 \mathbf{r}_{2} r 2 i i i p j p_{j} p j p 1 \mathbf{p}_{1} p 1 p 2 \mathbf{p}_{2} p 2 j j j
由于Poisson括号 { q ˉ i , q ˉ j } \{\bar{q}_{i},\bar{q}_{j}\} { q ˉ i , q ˉ j } { p ˉ i , p ˉ j } \{\bar{p}_{i},\bar{p}_{j}\} { p ˉ i , p ˉ j }
{ q ˉ i , q ˉ j } = 0 { p ˉ i , p ˉ j } = 0 \begin{aligned}
\{\bar{q}_{i},\bar{q}_{j}\}&=0\\
\{\bar{p}_{i},\bar{p}_{j}\}&=0
\end{aligned} { q ˉ i , q ˉ j } { p ˉ i , p ˉ j } = 0 = 0 其中 i i i j j j x x x y y y z z z { q ˉ i , p ˉ j } \{\bar{q}_{i},\bar{p}_{j}\} { q ˉ i , p ˉ j } { r i , p j } \{r_{i},p_{j}\} { r i , p j } { r C M i , p C M j } \{r_{CMi},p_{CMj}\} { r C M i , p C M j } { r C M i , p j } \{r_{CMi},p_{j}\} { r C M i , p j } { r i , p C M j } \{r_{i},p_{CMj}\} { r i , p C M j }
(1) { r i , p j } \{r_{i},p_{j}\} { r i , p j }
{ r i , p i } = ∑ α ( ∂ r i ∂ q α ∂ p i ∂ p α − ∂ r i ∂ p α ∂ p i ∂ q α ) = ∑ α ∂ r i ∂ q α ∂ p i ∂ p α = ∂ r i ∂ r 1 i ∂ p i ∂ p 1 i + ∂ r i ∂ r 2 i ∂ p i ∂ p 2 i = 1 ⋅ m 2 M + ( − 1 ) ⋅ ( − m 1 M ) = m 1 + m 2 M = 1 \begin{aligned}
\{r_{i},p_{i}\}&=\sum\limits_{\alpha}\left(\frac{\partial r_{i}}{\partial q_{\alpha}}\frac{\partial p_{i}}{\partial p_{\alpha}}-\frac{\partial r_{i}}{\partial p_{\alpha}}\frac{\partial p_{i}}{\partial q_{\alpha}}\right)\\
&=\sum\limits_{\alpha}\frac{\partial r_{i}}{\partial q_{\alpha}}\frac{\partial p_{i}}{\partial p_{\alpha}}\\
&=\frac{\partial r_{i}}{\partial r_{1i}}\frac{\partial p_{i}}{\partial p_{1i}}+\frac{\partial r_{i}}{\partial r_{2i}}\frac{\partial p_{i}}{\partial p_{2i}}\\
&=1\cdot \frac{m_{2}}{M}+(-1)\cdot \left(-\frac{m_{1}}{M}\right)\\
&=\frac{m_{1}+m_{2}}{M}\\
&=1
\end{aligned} { r i , p i } = α ∑ ( ∂ q α ∂ r i ∂ p α ∂ p i − ∂ p α ∂ r i ∂ q α ∂ p i ) = α ∑ ∂ q α ∂ r i ∂ p α ∂ p i = ∂ r 1 i ∂ r i ∂ p 1 i ∂ p i + ∂ r 2 i ∂ r i ∂ p 2 i ∂ p i = 1 ⋅ M m 2 + ( − 1 ) ⋅ ( − M m 1 ) = M m 1 + m 2 = 1 其中 q α q_{\alpha} q α p α p_{\alpha} p α 6 6 6 { r 1 x , r 1 y , r 1 z , r 2 x , r 2 y , r 2 z } \{r_{1x},r_{1y},r_{1z},r_{2x},r_{2y},r_{2z}\} { r 1 x , r 1 y , r 1 z , r 2 x , r 2 y , r 2 z } { r 1 i , r 2 i } \{r_{1i},r_{2i}\} { r 1 i , r 2 i } 6 6 6 { p 1 x , p 1 y , p 1 z , p 2 x , p 2 y , p 2 z } \{p_{1x},p_{1y},p_{1z},p_{2x},p_{2y},p_{2z}\} { p 1 x , p 1 y , p 1 z , p 2 x , p 2 y , p 2 z } { p 1 i , p 2 i } \{p_{1i},p_{2i}\} { p 1 i , p 2 i }
{ x i , y j } = ∑ α ( ∂ r i ∂ q α ∂ p j ∂ p α − ∂ r i ∂ p α ∂ p j ∂ q α ) = ∑ α ∂ r i ∂ q α ∂ p j ∂ p α = ∂ r i ∂ r 1 i ∂ p j ∂ p 1 i + ∂ r i ∂ r 2 i ∂ p j ∂ p 2 i + ∂ r i ∂ r 1 j ∂ p j ∂ p 1 j + ∂ r i ∂ r 2 j ∂ p j ∂ p 2 j = 1 ⋅ 0 + ( − 1 ) ⋅ 0 + 0 ⋅ m 2 M + 0 ⋅ ( − m 1 M ) = 0 \begin{aligned}
\{x_{i},y_{j}\}&=\sum\limits_{\alpha}\left(\frac{\partial r_{i}}{\partial q_{\alpha}}\frac{\partial p_{j}}{\partial p_{\alpha}}-\frac{\partial r_{i}}{\partial p_{\alpha}}\frac{\partial p_{j}}{\partial q_{\alpha}}\right)\\
&=\sum\limits_{\alpha}\frac{\partial r_{i}}{\partial q_{\alpha}}\frac{\partial p_{j}}{\partial p_{\alpha}}\\
&=\frac{\partial r_{i}}{\partial r_{1i}}\frac{\partial p_{j}}{\partial p_{1i}}+\frac{\partial r_{i}}{\partial r_{2i}}\frac{\partial p_{j}}{\partial p_{2i}}+\frac{\partial r_{i}}{\partial r_{1j}}\frac{\partial p_{j}}{\partial p_{1j}}+\frac{\partial r_{i}}{\partial r_{2j}}\frac{\partial p_{j}}{\partial p_{2j}}\\
&=1\cdot 0+(-1)\cdot 0+0\cdot \frac{m_{2}}{M}+0\cdot\left(-\frac{m_{1}}{M}\right)\\
&=0
\end{aligned} { x i , y j } = α ∑ ( ∂ q α ∂ r i ∂ p α ∂ p j − ∂ p α ∂ r i ∂ q α ∂ p j ) = α ∑ ∂ q α ∂ r i ∂ p α ∂ p j = ∂ r 1 i ∂ r i ∂ p 1 i ∂ p j + ∂ r 2 i ∂ r i ∂ p 2 i ∂ p j + ∂ r 1 j ∂ r i ∂ p 1 j ∂ p j + ∂ r 2 j ∂ r i ∂ p 2 j ∂ p j = 1 ⋅ 0 + ( − 1 ) ⋅ 0 + 0 ⋅ M m 2 + 0 ⋅ ( − M m 1 ) = 0 (2) { r C M i , p C M j } \{r_{CMi},p_{CMj}\} { r C M i , p C M j }
{ r C M i , p C M i } = ∑ α ( ∂ r C M i ∂ q α ∂ p C M i ∂ p α − ∂ r C M i ∂ p α ∂ p C M i ∂ q α ) = ∑ α ∂ r C M i ∂ q α ∂ p C M i ∂ p α = ∂ r C M i ∂ r 1 i ∂ p C M i ∂ p 1 i + ∂ r C M i ∂ r 2 i ∂ p C M i ∂ p 2 i = m 1 M ⋅ 1 + m 2 M ⋅ 1 = m 1 + m 2 M = 1 \begin{aligned}
\{r_{CMi},p_{CMi}\}&=\sum\limits_{\alpha}\left(\frac{\partial r_{CMi}}{\partial q_{\alpha}}\frac{\partial p_{CMi}}{\partial p_{\alpha}}-\frac{\partial r_{CMi}}{\partial p_{\alpha}}\frac{\partial p_{CMi}}{\partial q_{\alpha}}\right)\\
&=\sum\limits_{\alpha}\frac{\partial r_{CMi}}{\partial q_{\alpha}}\frac{\partial p_{CMi}}{\partial p_{\alpha}}\\
&=\frac{\partial r_{CMi}}{\partial r_{1i}}\frac{\partial p_{CMi}}{\partial p_{1i}}+\frac{\partial r_{CMi}}{\partial r_{2i}}\frac{\partial p_{CMi}}{\partial p_{2i}}\\
&=\frac{m_{1}}{M}\cdot 1+\frac{m_{2}}{M}\cdot 1\\
&=\frac{m_{1}+m_{2}}{M}\\
&=1
\end{aligned} { r C M i , p C M i } = α ∑ ( ∂ q α ∂ r C M i ∂ p α ∂ p C M i − ∂ p α ∂ r C M i ∂ q α ∂ p C M i ) = α ∑ ∂ q α ∂ r C M i ∂ p α ∂ p C M i = ∂ r 1 i ∂ r C M i ∂ p 1 i ∂ p C M i + ∂ r 2 i ∂ r C M i ∂ p 2 i ∂ p C M i = M m 1 ⋅ 1 + M m 2 ⋅ 1 = M m 1 + m 2 = 1 { r C M i , p C M j } = ∑ α ( ∂ r C M i ∂ q α ∂ p C M j ∂ p α − ∂ r C M i ∂ p α ∂ p C M j ∂ q α ) = ∑ α ∂ r C M i ∂ q α ∂ p C M j ∂ p α = ∂ r C M i ∂ r 1 i ∂ p C M j ∂ p 1 i + ∂ r C M i ∂ r 2 i ∂ p C M j ∂ p 2 i + ∂ r C M i ∂ r 1 j ∂ p C M j ∂ p 1 j + ∂ r C M i ∂ r 2 j ∂ p C M j ∂ p 2 j = 0 \begin{aligned}
\{r_{CMi},p_{CMj}\}&=\sum\limits_{\alpha}\left(\frac{\partial r_{CMi}}{\partial q_{\alpha}}\frac{\partial p_{CMj}}{\partial p_{\alpha}}-\frac{\partial r_{CMi}}{\partial p_{\alpha}}\frac{\partial p_{CMj}}{\partial q_{\alpha}}\right)\\
&=\sum\limits_{\alpha}\frac{\partial r_{CMi}}{\partial q_{\alpha}}\frac{\partial p_{CMj}}{\partial p_{\alpha}}\\
&=\frac{\partial r_{CMi}}{\partial r_{1i}}\frac{\partial p_{CMj}}{\partial p_{1i}}+\frac{\partial r_{CMi}}{\partial r_{2i}}\frac{\partial p_{CMj}}{\partial p_{2i}}+\frac{\partial r_{CMi}}{\partial r_{1j}}\frac{\partial p_{CMj}}{\partial p_{1j}}+\frac{\partial r_{CMi}}{\partial r_{2j}}\frac{\partial p_{CMj}}{\partial p_{2j}}\\
&=0
\end{aligned} { r C M i , p C M j } = α ∑ ( ∂ q α ∂ r C M i ∂ p α ∂ p C M j − ∂ p α ∂ r C M i ∂ q α ∂ p C M j ) = α ∑ ∂ q α ∂ r C M i ∂ p α ∂ p C M j = ∂ r 1 i ∂ r C M i ∂ p 1 i ∂ p C M j + ∂ r 2 i ∂ r C M i ∂ p 2 i ∂ p C M j + ∂ r 1 j ∂ r C M i ∂ p 1 j ∂ p C M j + ∂ r 2 j ∂ r C M i ∂ p 2 j ∂ p C M j = 0 (3) { r C M i , p j } \{r_{CMi},p_{j}\} { r C M i , p j } { r i , p C M j } \{r_{i},p_{CMj}\} { r i , p C M j }
{ r C M i , p j } = ∑ α ( ∂ r C M i ∂ q α ∂ p i ∂ p α − ∂ r C M i ∂ p α ∂ p i ∂ q α ) = ∑ α ∂ r C M i ∂ q α ∂ p i ∂ p α = ∂ r C M i ∂ r 1 i ∂ p i ∂ p 1 i + ∂ r C M i ∂ r 2 i ∂ p i ∂ p 2 i = m 1 M ⋅ m 2 M + m 2 M ⋅ ( − m 1 M ) = 0 \begin{aligned}
\{r_{CMi},p_{j}\}&=\sum\limits_{\alpha}\left(\frac{\partial r_{CMi}}{\partial q_{\alpha}}\frac{\partial p_{i}}{\partial p_{\alpha}}-\frac{\partial r_{CMi}}{\partial p_{\alpha}}\frac{\partial p_{i}}{\partial q_{\alpha}}\right)\\
&=\sum\limits_{\alpha}\frac{\partial r_{CMi}}{\partial q_{\alpha}}\frac{\partial p_{i}}{\partial p_{\alpha}}\\
&=\frac{\partial r_{CMi}}{\partial r_{1i}}\frac{\partial p_{i}}{\partial p_{1i}}+\frac{\partial r_{CMi}}{\partial r_{2i}}\frac{\partial p_{i}}{\partial p_{2i}}\\
&=\frac{m_{1}}{M}\cdot \frac{m_{2}}{M}+\frac{m_{2}}{M}\cdot \left(-\frac{m_{1}}{M}\right)\\
&=0
\end{aligned} { r C M i , p j } = α ∑ ( ∂ q α ∂ r C M i ∂ p α ∂ p i − ∂ p α ∂ r C M i ∂ q α ∂ p i ) = α ∑ ∂ q α ∂ r C M i ∂ p α ∂ p i = ∂ r 1 i ∂ r C M i ∂ p 1 i ∂ p i + ∂ r 2 i ∂ r C M i ∂ p 2 i ∂ p i = M m 1 ⋅ M m 2 + M m 2 ⋅ ( − M m 1 ) = 0 { r i , p C M j } = ∑ α ( ∂ r i ∂ q α ∂ p C M i ∂ p α − ∂ r i ∂ p α ∂ p C M i ∂ q α ) = ∑ α ∂ r i ∂ q α ∂ p C M i ∂ p α = ∂ r i ∂ r 1 i ∂ p C M i ∂ p 1 i + ∂ r i ∂ r 2 i ∂ p C M i ∂ p 2 i = 1 ⋅ 1 + ( − 1 ) ⋅ 1 = 0 \begin{aligned}
\{r_{i},p_{CMj}\}&=\sum\limits_{\alpha}\left(\frac{\partial r_{i}}{\partial q_{\alpha}}\frac{\partial p_{CMi}}{\partial p_{\alpha}}-\frac{\partial r_{i}}{\partial p_{\alpha}}\frac{\partial p_{CMi}}{\partial q_{\alpha}}\right)\\
&=\sum\limits_{\alpha}\frac{\partial r_{i}}{\partial q_{\alpha}}\frac{\partial p_{CMi}}{\partial p_{\alpha}}\\
&=\frac{\partial r_{i}}{\partial r_{1i}}\frac{\partial p_{CMi}}{\partial p_{1i}}+\frac{\partial r_{i}}{\partial r_{2i}}\frac{\partial p_{CMi}}{\partial p_{2i}}\\
&=1\cdot 1+(-1)\cdot 1\\
&=0
\end{aligned} { r i , p C M j } = α ∑ ( ∂ q α ∂ r i ∂ p α ∂ p C M i − ∂ p α ∂ r i ∂ q α ∂ p C M i ) = α ∑ ∂ q α ∂ r i ∂ p α ∂ p C M i = ∂ r 1 i ∂ r i ∂ p 1 i ∂ p C M i + ∂ r 2 i ∂ r i ∂ p 2 i ∂ p C M i = 1 ⋅ 1 + ( − 1 ) ⋅ 1 = 0 { r C M i , p j } = ∑ α ( ∂ r C M i ∂ q α ∂ p j ∂ p α − ∂ r C M i ∂ p α ∂ p j ∂ q α ) = ∑ α ∂ r C M i ∂ q α ∂ p j ∂ p α = ∂ r C M i ∂ r 1 i ∂ p j ∂ p 1 i + ∂ r C M i ∂ r 2 i ∂ p j ∂ p 2 i + ∂ r C M i ∂ r 1 j ∂ p j ∂ p 1 j + ∂ r C M i ∂ r 2 j ∂ p j ∂ p 2 j = 0 \begin{aligned}
\{r_{CMi},p_{j}\}&=\sum\limits_{\alpha}\left(\frac{\partial r_{CMi}}{\partial q_{\alpha}}\frac{\partial p_{j}}{\partial p_{\alpha}}-\frac{\partial r_{CMi}}{\partial p_{\alpha}}\frac{\partial p_{j}}{\partial q_{\alpha}}\right)\\
&=\sum\limits_{\alpha}\frac{\partial r_{CMi}}{\partial q_{\alpha}}\frac{\partial p_{j}}{\partial p_{\alpha}}\\
&=\frac{\partial r_{CMi}}{\partial r_{1i}}\frac{\partial p_{j}}{\partial p_{1i}}+\frac{\partial r_{CMi}}{\partial r_{2i}}\frac{\partial p_{j}}{\partial p_{2i}}+\frac{\partial r_{CMi}}{\partial r_{1j}}\frac{\partial p_{j}}{\partial p_{1j}}+\frac{\partial r_{CMi}}{\partial r_{2j}}\frac{\partial p_{j}}{\partial p_{2j}}\\
&=0
\end{aligned} { r C M i , p j } = α ∑ ( ∂ q α ∂ r C M i ∂ p α ∂ p j − ∂ p α ∂ r C M i ∂ q α ∂ p j ) = α ∑ ∂ q α ∂ r C M i ∂ p α ∂ p j = ∂ r 1 i ∂ r C M i ∂ p 1 i ∂ p j + ∂ r 2 i ∂ r C M i ∂ p 2 i ∂ p j + ∂ r 1 j ∂ r C M i ∂ p 1 j ∂ p j + ∂ r 2 j ∂ r C M i ∂ p 2 j ∂ p j = 0 { r i , p C M j } = ∑ α ( ∂ r i ∂ q α ∂ p C M j ∂ p α − ∂ r i ∂ p α ∂ p C M j ∂ q α ) = ∑ α ∂ r i ∂ q α ∂ p C M j ∂ p α = ∂ r i ∂ r 1 i ∂ p C M j ∂ p 1 i + ∂ r i ∂ r 2 i ∂ p C M j ∂ p 2 i + ∂ r i ∂ r 1 j ∂ p C M j ∂ p 1 j + ∂ r i ∂ r 2 j ∂ p C M j ∂ p 2 j = 0 \begin{aligned}
\{r_{i},p_{CMj}\}&=\sum\limits_{\alpha}\left(\frac{\partial r_{i}}{\partial q_{\alpha}}\frac{\partial p_{CMj}}{\partial p_{\alpha}}-\frac{\partial r_{i}}{\partial p_{\alpha}}\frac{\partial p_{CMj}}{\partial q_{\alpha}}\right)\\
&=\sum\limits_{\alpha}\frac{\partial r_{i}}{\partial q_{\alpha}}\frac{\partial p_{CMj}}{\partial p_{\alpha}}\\
&=\frac{\partial r_{i}}{\partial r_{1i}}\frac{\partial p_{CMj}}{\partial p_{1i}}+\frac{\partial r_{i}}{\partial r_{2i}}\frac{\partial p_{CMj}}{\partial p_{2i}}+\frac{\partial r_{i}}{\partial r_{1j}}\frac{\partial p_{CMj}}{\partial p_{1j}}+\frac{\partial r_{i}}{\partial r_{2j}}\frac{\partial p_{CMj}}{\partial p_{2j}}\\
&=0
\end{aligned} { r i , p C M j } = α ∑ ( ∂ q α ∂ r i ∂ p α ∂ p C M j − ∂ p α ∂ r i ∂ q α ∂ p C M j ) = α ∑ ∂ q α ∂ r i ∂ p α ∂ p C M j = ∂ r 1 i ∂ r i ∂ p 1 i ∂ p C M j + ∂ r 2 i ∂ r i ∂ p 2 i ∂ p C M j + ∂ r 1 j ∂ r i ∂ p 1 j ∂ p C M j + ∂ r 2 j ∂ r i ∂ p 2 j ∂ p C M j = 0 因此所有Poisson括号都满足要求，所以该变换是正则变换。
■ ~\tag*{$\blacksquare$} ■ 习题 2.7.7 验证
q ˉ = ln ( q − 1 sin p ) p ˉ = q cot p \begin{gathered}
\bar{q}=\ln \left(q^{-1} \sin p\right) \\
\bar{p}=q \cot p
\end{gathered} q ˉ = ln ( q − 1 sin p ) p ˉ = q cot p 是一个正则变换。
解答: 偏导数为
∂ q ˉ ∂ q = − q − 1 ∂ q ˉ ∂ p = cot p ∂ p ˉ ∂ q = cot p ∂ p ˉ ∂ p = − q ( 1 + cot 2 p ) \begin{aligned}
&\frac{\partial \bar{q}}{\partial q}=-q^{-1}&\qquad &\frac{\partial \bar{q}}{\partial p}=\cot p\\
&\frac{\partial \bar{p}}{\partial q}=\cot p&\qquad &\frac{\partial \bar{p}}{\partial p}=-q(1+\cot^{2} p)
\end{aligned} ∂ q ∂ q ˉ = − q − 1 ∂ q ∂ p ˉ = cot p ∂ p ∂ q ˉ = cot p ∂ p ∂ p ˉ = − q ( 1 + cot 2 p ) 唯一还需要验证的项是
{ q ˉ , p ˉ } = ∂ q ˉ ∂ q ∂ p ˉ ∂ p − ∂ q ˉ ∂ p ∂ p ˉ ∂ q = 1 \begin{aligned}
\{\bar{q},\bar{p}\}&=\frac{\partial \bar{q}}{\partial q}\frac{\partial \bar{p}}{\partial p}-\frac{\partial\bar{q}}{\partial p}\frac{\partial\bar{p}}{\partial q}\\
&=1
\end{aligned} { q ˉ , p ˉ } = ∂ q ∂ q ˉ ∂ p ∂ p ˉ − ∂ p ∂ q ˉ ∂ q ∂ p ˉ = 1 因此该变换是正则变换。
■ ~\tag*{$\blacksquare$} ■ 习题 2.7.8 我们希望在这里推导式 (2.7.9)，它给出了位形空间中坐标变换下动量的变换规律：
q i → q ˉ i ( q 1 , … , q n ) q_i \rightarrow \bar{q}_i(q_1, \ldots, q_n) q i → q ˉ i ( q 1 , … , q n ) (1) 说明若将上式反解为 q = q ( q ˉ ) q=q(\bar{q}) q = q ( q ˉ )
q ˙ i = ∑ j ∂ q i ∂ q ˉ j q ˉ ˙ j \dot{q}_i=\sum_j \frac{\partial q_i}{\partial \bar{q}_j} \dot{\bar{q}}_j q ˙ i = j ∑ ∂ q ˉ j ∂ q i q ˉ ˙ j (2) 由上式证明
( ∂ q ˙ i ∂ q ˙ j ) q ˉ = ∂ q i ∂ q ˉ j \left(\frac{\partial \dot{q}_i}{\partial \dot{q}_j}\right)_{\bar{q}}=\frac{\partial q_i}{\partial \bar{q}_j} ( ∂ q ˙ j ∂ q ˙ i ) q ˉ = ∂ q ˉ j ∂ q i (3) 现在计算
p ˉ i = [ ∂ L ( q ˉ , q ˉ ˙ ) ∂ q ˉ ˙ i ] q ˉ = [ ∂ L ( q , q ˙ ) ∂ q ˙ i ] q ˉ \bar{p}_i=\left[\frac{\partial \mathscr{L}(\bar{q}, \dot{\bar{q}})}{\partial \dot{\bar{q}}_i}\right]_{\bar{q}}=\left[\frac{\partial \mathscr{L}(q, \dot{q})}{\partial \dot{q}_i}\right]_{\bar{q}} p ˉ i = [ ∂ q ˉ ˙ i ∂ L ( q ˉ , q ˉ ˙ ) ] q ˉ = [ ∂ q ˙ i ∂ L ( q , q ˙ ) ] q ˉ 利用链式法则以及 q = q ( q ˉ ) q=q(\bar{q}) q = q ( q ˉ ) q ( q ˉ , q ˉ ˙ ) q(\bar{q}, \dot{\bar{q}}) q ( q ˉ , q ˉ ˙ )
(4) 通过计算式 (2.7.18) 中的Poisson括号，验证点变换是正则变换。
解答: (1) 由于 q i = q i ( q ˉ 1 , … , q ˉ n ) q_{i}=q_{i}(\bar{q}_{1},\ldots,\bar{q}_{n}) q i = q i ( q ˉ 1 , … , q ˉ n )
q ˙ i = d q i d t = ∑ j ∂ q i ∂ q ˉ j d q ˉ j d t = ∑ j ∂ q i ∂ q ˉ j q ˉ ˙ j \dot{q}_{i}=\frac{\mathrm{d}q_{i}}{\mathrm{d}t}=\sum_{j}\frac{\partial q_{i}}{\partial \bar{q}_{j}}\frac{\mathrm{d}\bar{q}_{j}}{\mathrm{d}t}=\sum_{j}\frac{\partial q_{i}}{\partial \bar{q}_{j}}\dot{\bar{q}}_{j} q ˙ i = d t d q i = j ∑ ∂ q ˉ j ∂ q i d t d q ˉ j = j ∑ ∂ q ˉ j ∂ q i q ˉ ˙ j (2) 由于速度 q ˉ ˙ j \dot{\bar{q}}_{j} q ˉ ˙ j q ˉ \bar{q} q ˉ
( ∂ q ˙ i ∂ q ˉ ˙ j ) q ˉ = ∂ ∂ q ˉ ˙ j ( ∑ l ∂ q i ∂ q ˉ k q ˉ ˙ k ) = ∑ k ∂ q i ∂ q ˉ k ∂ q ˉ ˙ k ∂ q ˉ ˙ j = ∑ k ∂ q i ∂ q ˉ k δ k j = ∂ q i ∂ q ˉ j (2.3) \left(\frac{\partial \dot{q}_{i}}{\partial \dot{\bar{q}}_{j}}\right)_{\bar{q}}=\frac{\partial}{\partial \dot{\bar{q}}_{j}}\left(\sum_{l}\frac{\partial q_{i}}{\partial \bar{q}_{k}}\dot{\bar{q}}_{k}
\right)=\sum_{k}\frac{\partial q_{i}}{\partial \bar{q}_{k}}\frac{\partial \dot{\bar{q}}_{k}}{\partial \dot{\bar{q}}_{j}}=\sum_{k}\frac{\partial q_{i}}{\partial \bar{q}_{k}}\delta_{kj}=\frac{\partial q_{i}}{\partial \bar{q}_{j}}\tag{2.3} ( ∂ q ˉ ˙ j ∂ q ˙ i ) q ˉ = ∂ q ˉ ˙ j ∂ ( l ∑ ∂ q ˉ k ∂ q i q ˉ ˙ k ) = k ∑ ∂ q ˉ k ∂ q i ∂ q ˉ ˙ j ∂ q ˉ ˙ k = k ∑ ∂ q ˉ k ∂ q i δ k j = ∂ q ˉ j ∂ q i ( 2.3 ) (3) 我们可以利用Lagrange量考察动量 p i p_{i} p i
p i : = ∂ L ∂ q ˙ i p_{i}:=\frac{\partial \mathscr{L}}{\partial \dot{q}_{i}} p i := ∂ q ˙ i ∂ L 若用新坐标和新速度表示Lagrange量，即 L = L ( q ˉ , q ˉ ˙ ) \mathscr{L}=\mathscr{L}(\bar{q},\dot{\bar{q}}) L = L ( q ˉ , q ˉ ˙ )
p ˉ i = ∂ L ( q ˉ , q ˉ ˙ ) ∂ q ˉ ˙ i \bar{p}_{i}=\frac{\partial\mathscr{L}(\bar{q},\dot{\bar{q}})}{\partial \dot{\bar{q}}_{i}} p ˉ i = ∂ q ˉ ˙ i ∂ L ( q ˉ , q ˉ ˙ ) 这里值得注意的是，尽管 L ( q ˉ , q ˉ ˙ ) \mathscr{L}(\bar{q},\dot{\bar{q}}) L ( q ˉ , q ˉ ˙ ) L ( q , q ˙ ) \mathscr{L}(q,\dot{q}) L ( q , q ˙ ) q q q ( q , q ˙ ) (q,\dot{q}) ( q , q ˙ ) q ˉ \bar{q} q ˉ ( q ˉ , q ˉ ˙ ) (\bar{q},\dot{\bar{q}}) ( q ˉ , q ˉ ˙ ) L ( q ˉ , q ˉ ˙ ) = L ( q , q ˙ ) \mathscr{L}(\bar{q},\dot{\bar{q}})=\mathscr{L}(q,\dot{q}) L ( q ˉ , q ˉ ˙ ) = L ( q , q ˙ )
p ˉ i = ( ∂ L ( q ˉ , q ˉ ˙ ) ∂ q ˉ ˙ i ) q ˉ = ( ∂ L ( q , q ˙ ) ∂ q ˉ ˙ i ) q ˉ \bar{p}_{i}=\left(\frac{\partial\mathscr{L}(\bar{q},\dot{\bar{q}})}{\partial \dot{\bar{q}}_{i}}\right)_{\bar{q}}=\left(\frac{\partial\mathscr{L}(q,\dot{q})}{\partial \dot{\bar{q}}_{i}}\right)_{\bar{q}} p ˉ i = ( ∂ q ˉ ˙ i ∂ L ( q ˉ , q ˉ ˙ ) ) q ˉ = ( ∂ q ˉ ˙ i ∂ L ( q , q ˙ ) ) q ˉ 也就是说，在保持 q ˉ \bar{q} q ˉ L \mathscr{L} L L \mathscr{L} L q ˉ ˙ i \dot{\bar{q}}_{i} q ˉ ˙ i
p ˉ i = ( ∂ L ( q , q ˙ ) ∂ q ˉ i ) q ˉ = ∑ j [ ∂ L ∂ q j ∂ q j ∂ q ˉ ˙ i + ∂ L ∂ q ˙ j ∂ q ˙ j ∂ q ˉ ˙ i ] \bar{p}_i=\left(\frac{\partial L(q, \dot{q})}{\partial \bar{q}_i}\right)_{\bar{q}}=\sum_j\left[\frac{\partial L}{\partial q_j} \frac{\partial q_j}{\partial \dot{\bar{q}}_i}+\frac{\partial L}{\partial \dot{q}_j} \frac{\partial \dot{q}_j}{\partial \dot{\bar{q}}_i}\right] p ˉ i = ( ∂ q ˉ i ∂ L ( q , q ˙ ) ) q ˉ = j ∑ [ ∂ q j ∂ L ∂ q ˉ ˙ i ∂ q j + ∂ q ˙ j ∂ L ∂ q ˉ ˙ i ∂ q ˙ j ] 由于坐标 q q q q ˉ ˙ \dot{\bar{q}} q ˉ ˙
p ˉ i = ∑ j ∂ L ∂ q ˙ j ∂ q ˙ j ∂ q ˉ ˙ i = ∑ j ∂ L ∂ q ˙ j ∂ q j ∂ q ˉ i = ∑ j ∂ q j ∂ q ˉ i p j \begin{aligned}
\bar{p}_i &=\sum_j\frac{\partial L}{\partial \dot{q}_j} \frac{\partial \dot{q}_j}{\partial \dot{\bar{q}}_i}\\
& =\sum_j \frac{\partial L}{\partial \dot{q}_j} \frac{\partial q_j}{\partial \bar{q}_i} \\
& =\sum_j \frac{\partial q_j}{\partial \bar{q}_i} p_j
\end{aligned} p ˉ i = j ∑ ∂ q ˙ j ∂ L ∂ q ˉ ˙ i ∂ q ˙ j = j ∑ ∂ q ˙ j ∂ L ∂ q ˉ i ∂ q j = j ∑ ∂ q ˉ i ∂ q j p j 最后一个等号使用了正则动量的定义。由此我们得到了式 (2.7.9)。
(4) 点变换为
q ˉ i = q ˉ i ( q 1 , … , q n ) p ˉ i = ∑ j ∂ q j ∂ q ˉ i p j \begin{aligned}
\bar{q}_{i}&=\bar{q}_{i}(q_{1},\ldots,q_{n})\\
\bar{p}_{i}&=\sum\limits_{j}\frac{\partial q_{j}}{\partial \bar{q}_{i}}p_{j}
\end{aligned} q ˉ i p ˉ i = q ˉ i ( q 1 , … , q n ) = j ∑ ∂ q ˉ i ∂ q j p j 在这种情况下，到 q ˉ \bar{q} q ˉ p ˉ i \bar{p}_i p ˉ i ∂ q i ∂ q ˉ j \frac{\partial q_i}{\partial \bar{q}_j} ∂ q ˉ j ∂ q i q ˉ \bar{q} q ˉ
{ q ˉ i , q ˉ j } = { p ˉ i , p ˉ j } = 0 \left\{\bar{q}_i, \bar{q}_j\right\}=\left\{\bar{p}_i, \bar{p}_j\right\}=0 { q ˉ i , q ˉ j } = { p ˉ i , p ˉ j } = 0 对于混合括号，有
{ q ˉ i , p ˉ j } = ∑ k ( ∂ q ˉ i ∂ q k ∂ p ˉ j ∂ p k − ∂ q ˉ i ∂ p k ∂ p ˉ j ∂ q k ) = ∑ k ∂ q ˉ i ∂ q k ( ∂ ∂ p k ( ∑ l ∂ q l ∂ q ˉ j p l ) ) = ∑ k ∂ q ˉ i ∂ q k ( ∑ l ∂ q l ∂ q ˉ j ∂ p l ∂ p k ) = ∑ k ∂ q ˉ i ∂ q k ( ∑ l ∂ q l ∂ q ˉ j δ l k ) = ∑ k ∂ q ˉ i ∂ q k ∂ q k ∂ q ˉ j = ∂ q ˉ i ∂ q ˉ j = δ i j \begin{aligned}
\left\{\bar{q}_i, \bar{p}_j\right\} & =\sum_k\left(\frac{\partial \bar{q}_i}{\partial q_k} \frac{\partial \bar{p}_j}{\partial p_k}-\frac{\partial \bar{q}_i}{\partial p_k} \frac{\partial \bar{p}_j}{\partial q_k}\right) \\
&=\sum_k \frac{\partial \bar{q}_i}{\partial q_k}
\left(\frac{\partial}{\partial p_{k}}\left(\sum_l \frac{\partial q_{l}}{\partial \bar{q}_{j}}p_{l}\right)\right)\\
&=\sum_k \frac{\partial \bar{q}_i}{\partial q_k}
\left(\sum_l \frac{\partial q_{l}}{\partial \bar{q}_{j}}\frac{\partial p_{l}}{\partial p_{k}}\right)\\
&=\sum_k \frac{\partial \bar{q}_i}{\partial q_k}
\left(\sum_l \frac{\partial q_{l}}{\partial \bar{q}_{j}}\delta_{lk}\right)\\
&=\sum_k \frac{\partial \bar{q}_i}{\partial q_k} \frac{\partial q_k}{\partial \bar{q}_j} \\
&=\frac{\partial \bar{q}_i}{\partial \bar{q}_j} \\
& =\delta_{i j}
\end{aligned} { q ˉ i , p ˉ j } = k ∑ ( ∂ q k ∂ q ˉ i ∂ p k ∂ p ˉ j − ∂ p k ∂ q ˉ i ∂ q k ∂ p ˉ j ) = k ∑ ∂ q k ∂ q ˉ i ( ∂ p k ∂ ( l ∑ ∂ q ˉ j ∂ q l p l ) ) = k ∑ ∂ q k ∂ q ˉ i ( l ∑ ∂ q ˉ j ∂ q l ∂ p k ∂ p l ) = k ∑ ∂ q k ∂ q ˉ i ( l ∑ ∂ q ˉ j ∂ q l δ l k ) = k ∑ ∂ q k ∂ q ˉ i ∂ q ˉ j ∂ q k = ∂ q ˉ j ∂ q ˉ i = δ ij 因此点变换是正则变换。
■ ~\tag*{$\blacksquare$} ■ 习题 2.7.9 直接计算验证式 (2.7.19)。利用链式法则将对 q , p q,p q , p q ˉ , p ˉ \bar{q},\bar{p} q ˉ , p ˉ 解答: 两个函数的Poisson括号定义为
{ ω , σ } = ∑ i ( ∂ ω ∂ q i ∂ σ ∂ p i − ∂ ω ∂ p i ∂ σ ∂ q i ) \{\omega, \sigma\}=\sum_i\left(\frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial p_i}-\frac{\partial \omega}{\partial p_i} \frac{\partial \sigma}{\partial q_i}\right) { ω , σ } = i ∑ ( ∂ q i ∂ ω ∂ p i ∂ σ − ∂ p i ∂ ω ∂ q i ∂ σ ) 计算Poisson括号需要知道在所采用的具体坐标系中，ω \omega ω σ \sigma σ q i q_i q i p i p_i p i
最简单的做法是把正则变换写成
q ˉ i = q ˉ i ( q , p ) p ˉ i = p ˉ ( q , p ) \begin{aligned}
\bar{q}_i & =\bar{q}_i(q, p) \\
\bar{p}_i & =\bar{p}(q, p)
\end{aligned} q ˉ i p ˉ i = q ˉ i ( q , p ) = p ˉ ( q , p ) 于是可以把新坐标中的Poisson括号写成
{ ω , σ } q ˉ , p ˉ = ∑ j ( ∂ ω ∂ q ˉ j ∂ σ ∂ p ˉ j − ∂ ω ∂ p ˉ j ∂ σ ∂ q ˉ j ) \{\omega, \sigma\}_{\bar{q}, \bar{p}}=\sum_j\left(\frac{\partial \omega}{\partial \bar{q}_j} \frac{\partial \sigma}{\partial \bar{p}_j}-\frac{\partial \omega}{\partial \bar{p}_j} \frac{\partial \sigma}{\partial \bar{q}_j}\right) { ω , σ } q ˉ , p ˉ = j ∑ ( ∂ q ˉ j ∂ ω ∂ p ˉ j ∂ σ − ∂ p ˉ j ∂ ω ∂ q ˉ j ∂ σ ) 假设该变换可逆，我们可以利用链式法则计算对带横线坐标的导数。得到如下结果（这里采用Einstein求和约定）：
{ ω , σ } q ˉ , p ˉ = ( ∂ ω ∂ q i ∂ q i ∂ q ˉ j + ∂ ω ∂ p i ∂ p i ∂ q ˉ j ) ( ∂ σ ∂ q k ∂ q k ∂ p ˉ j + ∂ σ ∂ p k ∂ p k ∂ p ˉ j ) − ( ∂ ω ∂ q i ∂ q i ∂ p ˉ j + ∂ ω ∂ p i ∂ p i ∂ p ˉ j ) ( ∂ σ ∂ q k ∂ q k ∂ q ˉ j + ∂ σ ∂ p k ∂ p k ∂ q ˉ j ) = ∂ ω ∂ q i ∂ σ ∂ p k ( ∂ q i ∂ q ˉ j ∂ p k ∂ p ˉ j − ∂ q i ∂ p ˉ j ∂ p k ∂ q ˉ j ) + ∂ ω ∂ p i ∂ σ ∂ q k ( ∂ p i ∂ q ˉ j ∂ q k ∂ p ˉ j − ∂ p i ∂ p ˉ j ∂ q k ∂ q ˉ j ) + ∂ ω ∂ q i ∂ σ ∂ q k ( ∂ q i ∂ q ˉ j ∂ q k ∂ p ˉ j − ∂ q i ∂ p ˉ j ∂ q k ∂ q ˉ j ) + ∂ ω ∂ p i ∂ σ ∂ p k ( ∂ p i ∂ q ˉ j ∂ p k ∂ p ˉ j − ∂ p i ∂ p ˉ j ∂ p k ∂ q ˉ j ) = ∂ ω ∂ q i ∂ σ ∂ p k { q i , p k } + ∂ ω ∂ p i ∂ σ ∂ q k { p i , q k } + ∂ ω ∂ q i ∂ σ ∂ q k { q i , q k } + ∂ ω ∂ p i ∂ σ ∂ p k { p i , p k } \begin{aligned}
\{\omega, \sigma\}_{\bar{q}, \bar{p}}=&\left(\frac{\partial \omega}{\partial q_i} \frac{\partial q_i}{\partial \bar{q}_j}+\frac{\partial \omega}{\partial p_i} \frac{\partial p_i}{\partial \bar{q}_j}\right)\left(\frac{\partial \sigma}{\partial q_k} \frac{\partial q_k}{\partial \bar{p}_j}+\frac{\partial \sigma}{\partial p_k} \frac{\partial p_k}{\partial \bar{p}_j}\right)\\
-&\left(\frac{\partial \omega}{\partial q_i} \frac{\partial q_i}{\partial \bar{p}_j}+\frac{\partial \omega}{\partial p_i} \frac{\partial p_i}{\partial \bar{p}_j}\right)\left(\frac{\partial \sigma}{\partial q_k} \frac{\partial q_k}{\partial \bar{q}_j}+\frac{\partial \sigma}{\partial p_k} \frac{\partial p_k}{\partial \bar{q}_j}\right) \\
=&\frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial p_k}\left(\frac{\partial q_i}{\partial \bar{q}_j} \frac{\partial p_k}{\partial \bar{p}_j}-\frac{\partial q_i}{\partial \bar{p}_j} \frac{\partial p_k}{\partial \bar{q}_j}\right)+\frac{\partial \omega}{\partial p_i} \frac{\partial \sigma}{\partial q_k}\left(\frac{\partial p_i}{\partial \bar{q}_j} \frac{\partial q_k}{\partial \bar{p}_j}-\frac{\partial p_i}{\partial \bar{p}_j} \frac{\partial q_k}{\partial \bar{q}_j}\right)\\
+&\frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial q_k}\left(\frac{\partial q_i}{\partial \bar{q}_j} \frac{\partial q_k}{\partial \bar{p}_j}-\frac{\partial q_i}{\partial \bar{p}_j} \frac{\partial q_k}{\partial \bar{q}_j}\right)+\frac{\partial \omega}{\partial p_i} \frac{\partial \sigma}{\partial p_k}\left(\frac{\partial p_i}{\partial \bar{q}_j} \frac{\partial p_k}{\partial \bar{p}_j}-\frac{\partial p_i}{\partial \bar{p}_j} \frac{\partial p_k}{\partial \bar{q}_j}\right) \\
=&\frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial p_k}\left\{q_i, p_k\right\}+\frac{\partial \omega}{\partial p_i} \frac{\partial \sigma}{\partial q_k}\left\{p_i, q_k\right\}+\frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial q_k}\left\{q_i, q_k\right\}+\frac{\partial \omega}{\partial p_i} \frac{\partial \sigma}{\partial p_k}\left\{p_i, p_k\right\}
\end{aligned} { ω , σ } q ˉ , p ˉ = − = + = ( ∂ q i ∂ ω ∂ q ˉ j ∂ q i + ∂ p i ∂ ω ∂ q ˉ j ∂ p i ) ( ∂ q k ∂ σ ∂ p ˉ j ∂ q k + ∂ p k ∂ σ ∂ p ˉ j ∂ p k ) ( ∂ q i ∂ ω ∂ p ˉ j ∂ q i + ∂ p i ∂ ω ∂ p ˉ j ∂ p i ) ( ∂ q k ∂ σ ∂ q ˉ j ∂ q k + ∂ p k ∂ σ ∂ q ˉ j ∂ p k ) ∂ q i ∂ ω ∂ p k ∂ σ ( ∂ q ˉ j ∂ q i ∂ p ˉ j ∂ p k − ∂ p ˉ j ∂ q i ∂ q ˉ j ∂ p k ) + ∂ p i ∂ ω ∂ q k ∂ σ ( ∂ q ˉ j ∂ p i ∂ p ˉ j ∂ q k − ∂ p ˉ j ∂ p i ∂ q ˉ j ∂ q k ) ∂ q i ∂ ω ∂ q k ∂ σ ( ∂ q ˉ j ∂ q i ∂ p ˉ j ∂ q k − ∂ p ˉ j ∂ q i ∂ q ˉ j ∂ q k ) + ∂ p i ∂ ω ∂ p k ∂ σ ( ∂ q ˉ j ∂ p i ∂ p ˉ j ∂ p k − ∂ p ˉ j ∂ p i ∂ q ˉ j ∂ p k ) ∂ q i ∂ ω ∂ p k ∂ σ { q i , p k } + ∂ p i ∂ ω ∂ q k ∂ σ { p i , q k } + ∂ q i ∂ ω ∂ q k ∂ σ { q i , q k } + ∂ p i ∂ ω ∂ p k ∂ σ { p i , p k } 对于正则变换，上一式中的Poisson括号满足
{ q i , p k } = − { p i , q k } = δ i k { q i , q k } = { p i , p k } = 0 \begin{array}{l}
\left\{q_i, p_k\right\}=-\left\{p_i, q_k\right\}=\delta_{i k} \\
\left\{q_i, q_k\right\}=\left\{p_i, p_k\right\}=0
\end{array} { q i , p k } = − { p i , q k } = δ ik { q i , q k } = { p i , p k } = 0 将这些条件代入上式，得到
{ ω , σ } q ˉ , p ˉ = ( ∂ ω ∂ q i ∂ σ ∂ p k − ∂ ω ∂ p i ∂ σ ∂ q k ) δ i k = ∂ ω ∂ q i ∂ σ ∂ p i − ∂ ω ∂ p i ∂ σ ∂ q i = { ω , σ } q , p \begin{aligned}
\{\omega, \sigma\}_{\bar{q}, \bar{p}} & =\left(\frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial p_k}-\frac{\partial \omega}{\partial p_i} \frac{\partial \sigma}{\partial q_k}\right) \delta_{i k} \\
& =\frac{\partial \omega}{\partial q_i} \frac{\partial \sigma}{\partial p_i}-\frac{\partial \omega}{\partial p_i} \frac{\partial \sigma}{\partial q_i} \\
& =\{\omega, \sigma\}_{q, p}
\end{aligned} { ω , σ } q ˉ , p ˉ = ( ∂ q i ∂ ω ∂ p k ∂ σ − ∂ p i ∂ ω ∂ q k ∂ σ ) δ ik = ∂ q i ∂ ω ∂ p i ∂ σ − ∂ p i ∂ ω ∂ q i ∂ σ = { ω , σ } q , p 因此Poisson括号在正则变换下不变。
■ ~\tag*{$\blacksquare$} ■ 2.8 对称性及其推论
习题 2.8.1 证明总动量 p = p 1 + p 2 p=p_1+p_2 p = p 1 + p 2
解答: 由于 g = p 1 + p 2 g=p_{1}+p_{2} g = p 1 + p 2
δ x 1 = + ε ∂ g ∂ p 1 = + ε , δ p 1 = − ε ∂ g ∂ x 1 = 0 δ x 2 = + ε ∂ g ∂ p 2 = + ε , δ p 2 = − ε ∂ g ∂ x 2 = 0 \begin{array}{ll}
\delta x_1=+\varepsilon \frac{\partial g}{\partial p_1}=+\varepsilon, & \delta p_1=-\varepsilon \frac{\partial g}{\partial x_1}=0 \\
\delta x_2=+\varepsilon \frac{\partial g}{\partial p_2}=+\varepsilon, & \delta p_2=-\varepsilon \frac{\partial g}{\partial x_2}=0
\end{array} δ x 1 = + ε ∂ p 1 ∂ g = + ε , δ x 2 = + ε ∂ p 2 ∂ g = + ε , δ p 1 = − ε ∂ x 1 ∂ g = 0 δ p 2 = − ε ∂ x 2 ∂ g = 0 因此保留到 ε \varepsilon ε x i → x ˉ i ( x j , p j ) x_{i}\to \bar{x}_{i}(x_{j},p_{j}) x i → x ˉ i ( x j , p j ) p i → p ˉ i ( x j , p j ) p_{i}\to \bar{p}_{i}(x_{j},p_{j}) p i → p ˉ i ( x j , p j )
x ˉ 1 = x 1 + ε , p ˉ 1 = p 1 , x ˉ 2 = x 2 + ε , p ˉ 2 = p 2 , \begin{array}{ll}
\bar{x}_1=x_1+\varepsilon, & \bar{p}_1=p_1, \\
\bar{x}_2=x_2+\varepsilon, & \bar{p}_2=p_2,
\end{array} x ˉ 1 = x 1 + ε , x ˉ 2 = x 2 + ε , p ˉ 1 = p 1 , p ˉ 2 = p 2 , 这恰好对应于整个系统平移 ε \varepsilon ε
■ ~\tag*{$\blacksquare$} ■ 习题 2.8.2 验证由任意动力学变量 g g g ε \varepsilon ε
解答: 若由动力学变量 g g g
q ˉ i = q i + ε ∂ g ∂ p i p ˉ j = p j − ε ∂ g ∂ q j \begin{aligned}
\bar{q}_{i}&=q_{i}+\varepsilon\frac{\partial g}{\partial p_{i}}\\
\bar{p}_{j}&=p_{j}-\varepsilon\frac{\partial g}{\partial q_{j}}
\end{aligned} q ˉ i p ˉ j = q i + ε ∂ p i ∂ g = p j − ε ∂ q j ∂ g 那么新坐标和新动量之间的Poisson括号为
{ q ˉ i , p ˉ j } = ∑ k ( ∂ q ˉ i ∂ q k ∂ p ˉ j ∂ p k − ∂ q ˉ i ∂ p k ∂ p ˉ j ∂ q k ) = ∑ k [ ( δ i k + ε ∂ 2 g ∂ p i ∂ q k ) ( δ j k + ε ∂ 2 g ∂ q j ∂ p k ) − ε ∂ 2 g ∂ p i ∂ p k ⋅ ε ∂ 2 g ∂ q i ∂ q k ] = ∑ k [ δ i k δ j k + ε ∂ 2 g ∂ p i ∂ q k ⋅ δ j k − δ i k ⋅ ε ∂ 2 g ∂ q j ∂ p k + O ( ε 2 ) ] = δ i j + ε ∂ 2 g ∂ p i ∂ q j − ε ∂ 2 g ∂ q j ∂ p i + O ( ε 2 ) = δ i j + O ( ε 2 ) ≈ δ i j \begin{aligned}
\{\bar{q}_{i},\bar{p}_{j}\}&=\sum\limits_{k}\left(\frac{\partial \bar{q}_{i}}{\partial q_{k}}\frac{\partial \bar{p}_{j}}{\partial p_{k}}-\frac{\partial \bar{q}_{i}}{\partial p_{k}}\frac{\partial \bar{p}_{j}}{\partial q_{k}}\right)\\
&=\sum\limits_{k}\left[\left(\delta_{ik}+\varepsilon\frac{\partial^{2}g}{\partial p_{i}\partial q_{k}}\right)\left(\delta_{jk}+\varepsilon\frac{\partial^{2}g}{\partial q_{j}\partial p_{k}}\right)-\varepsilon\frac{\partial^{2}g}{\partial p_{i}\partial p_{k}}\cdot\varepsilon\frac{\partial^{2}g}{\partial q_{i}\partial q_{k}}\right]\\
&=\sum\limits_{k}\left[\delta_{ik}\delta_{jk}+\varepsilon\frac{\partial^{2}g}{\partial p_{i}\partial q_{k}}\cdot\delta_{jk}-\delta_{ik}\cdot\varepsilon\frac{\partial^{2}g}{\partial q_{j}\partial p_{k}}+\mathcal{O}(\varepsilon^{2})\right]\\
&=\delta_{ij}+\varepsilon\frac{\partial^{2}g}{\partial p_{i}\partial q_{j}}-\varepsilon\frac{\partial^{2}g}{\partial q_{j}\partial p_{i}}+\mathcal{O}(\varepsilon^{2})\\
&=\delta_{ij}+\mathcal{O}(\varepsilon^{2})\\
&\approx \delta_{ij}
\end{aligned} { q ˉ i , p ˉ j } = k ∑ ( ∂ q k ∂ q ˉ i ∂ p k ∂ p ˉ j − ∂ p k ∂ q ˉ i ∂ q k ∂ p ˉ j ) = k ∑ [ ( δ ik + ε ∂ p i ∂ q k ∂ 2 g ) ( δ j k + ε ∂ q j ∂ p k ∂ 2 g ) − ε ∂ p i ∂ p k ∂ 2 g ⋅ ε ∂ q i ∂ q k ∂ 2 g ] = k ∑ [ δ ik δ j k + ε ∂ p i ∂ q k ∂ 2 g ⋅ δ j k − δ ik ⋅ ε ∂ q j ∂ p k ∂ 2 g + O ( ε 2 ) ] = δ ij + ε ∂ p i ∂ q j ∂ 2 g − ε ∂ q j ∂ p i ∂ 2 g + O ( ε 2 ) = δ ij + O ( ε 2 ) ≈ δ ij 因此，由任意动力学变量 g g g
■ ~\tag*{$\blacksquare$} ■ 习题 2.8.3 考虑
H = p x 2 + p y 2 2 m + 1 2 m ω 2 ( x 2 + y 2 ) \mathscr{H}=\frac{p_x^2+p_y^2}{2 m}+\frac{1}{2} m \omega^2\left(x^2+y^2\right) H = 2 m p x 2 + p y 2 + 2 1 m ω 2 ( x 2 + y 2 ) 它在坐标和动量同时旋转下的不变性导致 l z l_z l z H \mathscr{H} H g g g δ H \delta \mathscr{H} δ H ε { H , g } \varepsilon\{\mathscr{H}, g\} ε { H , g }
解答: 只旋转坐标：
{ x ˉ = x cos θ − y sin θ y ˉ = x sin θ + y cos θ { p ˉ x = p x p ˉ y = p y \left\{
\begin{aligned}
\bar{x}=x\cos\theta-y\sin\theta\\
\bar{y}=x\sin\theta+y\cos\theta
\end{aligned}
\right.
\qquad
\left\{
\begin{aligned}
\bar{p}_{x}=p_{x}\\
\bar{p}_{y}=p_{y}
\end{aligned}
\right. { x ˉ = x cos θ − y sin θ y ˉ = x sin θ + y cos θ { p ˉ x = p x p ˉ y = p y 于是Poisson括号为
{ x ˉ , y ˉ } = ∂ x ˉ ∂ x ∂ y ˉ ∂ p x − ∂ x ˉ ∂ p x ∂ y ˉ ∂ x + ∂ x ˉ ∂ y ∂ y ˉ ∂ p y − ∂ x ˉ ∂ p y ∂ y ˉ ∂ y = 0 { p ˉ x , p ˉ y } = { p x , p y } = 0 { x ˉ , p ˉ x } = ∂ x ˉ ∂ x ∂ p ˉ x ∂ p x − ∂ x ˉ ∂ p x ∂ p ˉ x ∂ x + ∂ x ˉ ∂ y ∂ p ˉ x ∂ p y − ∂ x ˉ ∂ p y ∂ p ˉ x ∂ y = cos θ ≠ 1 { x ˉ , p ˉ y } = ∂ x ˉ ∂ x ∂ p ˉ y ∂ p x − ∂ x ˉ ∂ p x ∂ p ˉ y ∂ x + ∂ x ˉ ∂ y ∂ p ˉ y ∂ p y − ∂ x ˉ ∂ p y ∂ p ˉ y ∂ y = − sin θ ≠ 0 { y ˉ , p ˉ x } = ∂ y ˉ ∂ x ∂ p ˉ x ∂ p x − ∂ y ˉ ∂ p x ∂ p ˉ x ∂ x + ∂ y ˉ ∂ y ∂ p ˉ x ∂ p y − ∂ y ˉ ∂ p y ∂ p ˉ x ∂ y = sin θ ≠ 0 { y ˉ , p ˉ y } = ∂ y ˉ ∂ x ∂ p ˉ y ∂ p x − ∂ y ˉ ∂ p x ∂ p ˉ y ∂ x + ∂ y ˉ ∂ y ∂ p ˉ y ∂ p y − ∂ y ˉ ∂ p y ∂ p ˉ y ∂ y = cos θ ≠ 1 \begin{aligned}
\{\bar{x},\bar{y}\}&=\frac{\partial \bar{x}}{\partial x}\frac{\partial \bar{y}}{\partial p_{x}}-\frac{\partial \bar{x}}{\partial p_{x}}\frac{\partial \bar{y}}{\partial x}+\frac{\partial \bar{x}}{\partial y}\frac{\partial \bar{y}}{\partial p_{y}}-\frac{\partial \bar{x}}{\partial p_{y}}\frac{\partial \bar{y}}{\partial y}=0\\
\{\bar{p}_{x},\bar{p}_{y}\}&=\{p_{x},p_{y}\}=0\\
\{\bar{x},\bar{p}_{x}\}&=\frac{\partial \bar{x}}{\partial x}\frac{\partial \bar{p}_{x}}{\partial p_{x}}-\frac{\partial \bar{x}}{\partial p_{x}}\frac{\partial \bar{p}_{x}}{\partial x}+\frac{\partial \bar{x}}{\partial y}\frac{\partial \bar{p}_{x}}{\partial p_{y}}-\frac{\partial \bar{x}}{\partial p_{y}}\frac{\partial \bar{p}_{x}}{\partial y}=\cos\theta\neq 1\\
\{\bar{x},\bar{p}_{y}\}&=\frac{\partial \bar{x}}{\partial x}\frac{\partial \bar{p}_{y}}{\partial p_{x}}-\frac{\partial \bar{x}}{\partial p_{x}}\frac{\partial \bar{p}_{y}}{\partial x}+\frac{\partial \bar{x}}{\partial y}\frac{\partial \bar{p}_{y}}{\partial p_{y}}-\frac{\partial \bar{x}}{\partial p_{y}}\frac{\partial \bar{p}_{y}}{\partial y}=-\sin\theta\neq 0\\
\{\bar{y},\bar{p}_{x}\}&=\frac{\partial \bar{y}}{\partial x}\frac{\partial \bar{p}_{x}}{\partial p_{x}}-\frac{\partial \bar{y}}{\partial p_{x}}\frac{\partial \bar{p}_{x}}{\partial x}+\frac{\partial \bar{y}}{\partial y}\frac{\partial \bar{p}_{x}}{\partial p_{y}}-\frac{\partial \bar{y}}{\partial p_{y}}\frac{\partial \bar{p}_{x}}{\partial y}=\sin\theta\neq 0\\
\{\bar{y},\bar{p}_{y}\}&=\frac{\partial \bar{y}}{\partial x}\frac{\partial \bar{p}_{y}}{\partial p_{x}}-\frac{\partial \bar{y}}{\partial p_{x}}\frac{\partial \bar{p}_{y}}{\partial x}+\frac{\partial \bar{y}}{\partial y}\frac{\partial \bar{p}_{y}}{\partial p_{y}}-\frac{\partial \bar{y}}{\partial p_{y}}\frac{\partial \bar{p}_{y}}{\partial y}=\cos\theta\neq 1\\
\end{aligned} { x ˉ , y ˉ } { p ˉ x , p ˉ y } { x ˉ , p ˉ x } { x ˉ , p ˉ y } { y ˉ , p ˉ x } { y ˉ , p ˉ y } = ∂ x ∂ x ˉ ∂ p x ∂ y ˉ − ∂ p x ∂ x ˉ ∂ x ∂ y ˉ + ∂ y ∂ x ˉ ∂ p y ∂ y ˉ − ∂ p y ∂ x ˉ ∂ y ∂ y ˉ = 0 = { p x , p y } = 0 = ∂ x ∂ x ˉ ∂ p x ∂ p ˉ x − ∂ p x ∂ x ˉ ∂ x ∂ p ˉ x + ∂ y ∂ x ˉ ∂ p y ∂ p ˉ x − ∂ p y ∂ x ˉ ∂ y ∂ p ˉ x = cos θ = 1 = ∂ x ∂ x ˉ ∂ p x ∂ p ˉ y − ∂ p x ∂ x ˉ ∂ x ∂ p ˉ y + ∂ y ∂ x ˉ ∂ p y ∂ p ˉ y − ∂ p y ∂ x ˉ ∂ y ∂ p ˉ y = − sin θ = 0 = ∂ x ∂ y ˉ ∂ p x ∂ p ˉ x − ∂ p x ∂ y ˉ ∂ x ∂ p ˉ x + ∂ y ∂ y ˉ ∂ p y ∂ p ˉ x − ∂ p y ∂ y ˉ ∂ y ∂ p ˉ x = sin θ = 0 = ∂ x ∂ y ˉ ∂ p x ∂ p ˉ y − ∂ p x ∂ y ˉ ∂ x ∂ p ˉ y + ∂ y ∂ y ˉ ∂ p y ∂ p ˉ y − ∂ p y ∂ y ˉ ∂ y ∂ p ˉ y = cos θ = 1 因此，只旋转坐标不是正则变换。
若 δ H = ε { H , g } = ε ( ∂ H ∂ x ∂ g ∂ p x − ∂ H ∂ p x ∂ g ∂ x + ∂ H ∂ y ∂ g ∂ p y − ∂ H ∂ p y ∂ g ∂ y ) \delta \mathscr{H}=\varepsilon\{\mathscr{H},g\}=\varepsilon\left(\frac{\partial \mathscr{H}}{\partial x}\frac{\partial g}{\partial p_{x}}-\frac{\partial \mathscr{H}}{\partial p_{x}}\frac{\partial g}{\partial x}+\frac{\partial \mathscr{H}}{\partial y}\frac{\partial g}{\partial p_{y}}-\frac{\partial \mathscr{H}}{\partial p_{y}}\frac{\partial g}{\partial y}\right) δ H = ε { H , g } = ε ( ∂ x ∂ H ∂ p x ∂ g − ∂ p x ∂ H ∂ x ∂ g + ∂ y ∂ H ∂ p y ∂ g − ∂ p y ∂ H ∂ y ∂ g )
δ x = ε ∂ g ∂ p x δ p x = − ε ∂ g ∂ x , δ y = ε ∂ g ∂ p y δ p y = − ε ∂ g ∂ y . \begin{array}{lll}
&\delta x=\varepsilon\dfrac{\partial g}{\partial p_{x}}\qquad &\delta p_{x}=-\varepsilon\dfrac{\partial g}{\partial x},\\
&\delta y=\varepsilon\dfrac{\partial g}{\partial p_{y}}\qquad &\delta p_{y}=-\varepsilon\dfrac{\partial g}{\partial y}.
\end{array} δ x = ε ∂ p x ∂ g δ y = ε ∂ p y ∂ g δ p x = − ε ∂ x ∂ g , δ p y = − ε ∂ y ∂ g . 这意味着
{ x ˉ = x + ε ∂ g ∂ p x y ˉ = y + ε ∂ g ∂ p y { p ˉ x = p x − ε ∂ g ∂ x p ˉ y = p y − ε ∂ g ∂ y \left\{\begin{aligned}
\bar{x}&=x+\varepsilon\frac{\partial g}{\partial p_{x}}\\
\bar{y}&=y+\varepsilon\frac{\partial g}{\partial p_{y}}
\end{aligned}\right.
\qquad
\left\{\begin{aligned}
\bar{p}_{x}&=p_{x}-\varepsilon\frac{\partial g}{\partial x}\\
\bar{p}_{y}&=p_{y}-\varepsilon\frac{\partial g}{\partial y}
\end{aligned}\right. ⎩ ⎨ ⎧ x ˉ y ˉ = x + ε ∂ p x ∂ g = y + ε ∂ p y ∂ g ⎩ ⎨ ⎧ p ˉ x p ˉ y = p x − ε ∂ x ∂ g = p y − ε ∂ y ∂ g 根据上一题，这是一个正则变换。因此，不存在任何 g g g δ H = ε { H , g } \delta \mathscr{H}=\varepsilon\{\mathscr{H},g\} δ H = ε { H , g }
■ ~\tag*{$\blacksquare$} ■ 习题 2.8.4 考虑 H = 1 2 p 2 + 1 2 x 2 \mathscr{H}=\frac{1}{2} p^2+\frac{1}{2} x^2 H = 2 1 p 2 + 2 1 x 2 x − p x-p x − p
解答: 考虑一个一维系统
H = 1 2 ( p 2 + x 2 ) \mathscr{H}=\frac{1}{2}(p^{2}+x^{2}) H = 2 1 ( p 2 + x 2 ) 并在相空间的 x − p x-p x − p
δ x = ε p δ p = − ε x \begin{aligned}
\delta x&=\varepsilon p\\
\delta p&=-\varepsilon x
\end{aligned} δ x δ p = εp = − ε x 这是一个正则变换，因为
{ x ˉ , p ˉ } = { x , p } + ε { δ x , p } + ε { x , δ p } + O ( ε 2 ) = { x , p } = 1 \begin{aligned}
\{\bar{x},\bar{p}\}&=\{x,p\}+\varepsilon\{\delta x,p\}+\varepsilon\{x,\delta p\}+\mathcal{O}(\varepsilon^{2})\\
&=\{x,p\}\\
&=1
\end{aligned} { x ˉ , p ˉ } = { x , p } + ε { δ x , p } + ε { x , δ p } + O ( ε 2 ) = { x , p } = 1 若 g ( x , p ) g(x,p) g ( x , p )
δ x = ε { x , g } = ε ∂ g ∂ p = ε p ⇒ ∂ g ∂ p = p δ p = ε { p , g } = − ε ∂ g ∂ x = − ε x ⇒ ∂ g ∂ x = x \begin{aligned}
\delta x&=\varepsilon\{x,g\}=\varepsilon\frac{\partial g}{\partial p}=\varepsilon p \Rightarrow \frac{\partial g}{\partial p}=p\\
\delta p&=\varepsilon\{p,g\}=-\varepsilon\frac{\partial g}{\partial x}=-\varepsilon x\Rightarrow\frac{\partial g}{\partial x}=x
\end{aligned} δ x δ p = ε { x , g } = ε ∂ p ∂ g = εp ⇒ ∂ p ∂ g = p = ε { p , g } = − ε ∂ x ∂ g = − ε x ⇒ ∂ x ∂ g = x 这两个方程的解为
g ( x , p ) = 1 2 ( p 2 + x 2 ) + C g(x,p)=\frac{1}{2}(p^{2}+x^{2})+C g ( x , p ) = 2 1 ( p 2 + x 2 ) + C 其中 C C C
事实上，这个正则变换就是取 θ = t \theta=t θ = t
■ ~\tag*{$\blacksquare$} ■ 习题 2.8.5 为什么一个使 H \mathscr{H} H H \mathscr{H} H ( x = a , y = 0 ) (x=a, y=0) ( x = a , y = 0 ) ( p x = b , p y = 0 ) \left(p_x=b, p_y=0\right) ( p x = b , p y = 0 ) ( x = 0 , y = a ) (x=0, y=a) ( x = 0 , y = a ) ( p x = b , p y = 0 ) \left(p_x=b, p_y=0\right) ( p x = b , p y = 0 ) H \mathscr{H} H
解答: 如果Hamilton量在一个常规正则变换下保持不变，并且我们能找到生成元 g g g
q ˉ i = q i + ε ∂ g ∂ p i ≡ q i + δ q i p ˉ i = p i − ε ∂ g ∂ q i ≡ p i + δ p i \begin{aligned}
\bar{q}_i & =q_i+\varepsilon \frac{\partial g}{\partial p_i} \equiv q_i+\delta q_i \\
\bar{p}_i & =p_i-\varepsilon \frac{\partial g}{\partial q_i} \equiv p_i+\delta p_i
\end{aligned} q ˉ i p ˉ i = q i + ε ∂ p i ∂ g ≡ q i + δ q i = p i − ε ∂ q i ∂ g ≡ p i + δ p i 那么 g g g
如果我们考虑一个有限的正则变换 ( q , p ) → ( q ˉ , p ˉ ) (q, p) \rightarrow(\bar{q}, \bar{p}) ( q , p ) → ( q ˉ , p ˉ ) ( q ( t ) , p ( t ) ) (q(t), p(t)) ( q ( t ) , p ( t ))
∂ H ∂ p i = q ˙ i − ∂ H ∂ q i = p ˙ i \begin{aligned}
\frac{\partial H}{\partial p_i} & =\dot{q}_i \\
-\frac{\partial H}{\partial q_i} & =\dot{p}_i
\end{aligned} ∂ p i ∂ H − ∂ q i ∂ H = q ˙ i = p ˙ i 则把原轨迹中的每一点 ( q ( t ) , p ( t ) ) (q(t), p(t)) ( q ( t ) , p ( t )) ( q ˉ ( t ) , p ˉ ( t ) ) (\bar{q}(t), \bar{p}(t)) ( q ˉ ( t ) , p ˉ ( t ))
∂ H ∂ p ˉ i = q ˉ ˙ i − ∂ H ∂ q ˉ i = p ˉ ˙ i \begin{align}
\frac{\partial H}{\partial \bar{p}_i} & =\dot{\bar{q}}_i \tag{2.4}\\
-\frac{\partial H}{\partial \bar{q}_i} & =\dot{\bar{p}}_i\tag{2.5}
\end{align} ∂ p ˉ i ∂ H − ∂ q ˉ i ∂ H = q ˉ ˙ i = p ˉ ˙ i ( 2.4 ) ( 2.5 ) 这个证明稍微有些微妙，过程如下。首先回顾正则变换条件的推导。该推导针对的是被动变换，其中两组参数 ( q , p ) → ( q ˉ , p ˉ ) (q, p) \rightarrow(\bar{q}, \bar{p}) ( q , p ) → ( q ˉ , p ˉ ) ( q , p ) → ( q ˉ , p ˉ ) (q, p) \rightarrow(\bar{q}, \bar{p}) ( q , p ) → ( q ˉ , p ˉ ) ( q , p ) (q, p) ( q , p ) ( q ˉ , p ˉ ) (\bar{q}, \bar{p}) ( q ˉ , p ˉ ) H ( q ˉ , p ˉ ) = H ( q , p ) H(\bar{q}, \bar{p})=H(q, p) H ( q ˉ , p ˉ ) = H ( q , p ) ( q , p ) (q, p) ( q , p ) ( q ˉ , p ˉ ) (\bar{q}, \bar{p}) ( q ˉ , p ˉ ) H ( q ˉ , p ˉ ) = H ( q , p ) H(\bar{q}, \bar{p})=H(q, p) H ( q ˉ , p ˉ ) = H ( q , p )
q ˉ ˙ j = ∑ k ∂ H ∂ q ˉ k { q ˉ j , q ˉ k } + ∑ k ∂ H ∂ p ˉ k { q ˉ j , p ˉ k } p ˉ ˙ j = ∑ k ∂ H ∂ q ˉ k { p ˉ j , q ˉ k } + ∑ k ∂ H ∂ p ˉ k { p ˉ j , p ˉ k } \begin{align}
\dot{\bar{q}}_j&=\sum_k \frac{\partial H}{\partial \bar{q}_k}\left\{\bar{q}_j, \bar{q}_k\right\}+\sum_k \frac{\partial H}{\partial \bar{p}_k}\left\{\bar{q}_j, \bar{p}_k\right\} \tag{2.6}\\
\dot{\bar{p}}_j&=\sum_k \frac{\partial H}{\partial \bar{q}_k}\left\{\bar{p}_j, \bar{q}_k\right\}+\sum_k \frac{\partial H}{\partial \bar{p}_k}\left\{\bar{p}_j, \bar{p}_k\right\}\tag{2.7}
\end{align} q ˉ ˙ j p ˉ ˙ j = k ∑ ∂ q ˉ k ∂ H { q ˉ j , q ˉ k } + k ∑ ∂ p ˉ k ∂ H { q ˉ j , p ˉ k } = k ∑ ∂ q ˉ k ∂ H { p ˉ j , q ˉ k } + k ∑ ∂ p ˉ k ∂ H { p ˉ j , p ˉ k } ( 2.6 ) ( 2.7 ) 由于该变换被指定为正则变换，所以这里Poisson括号满足相应条件：
{ q ˉ j , q ˉ k } = { p ˉ j , p ˉ k } = 0 { q ˉ j , p ˉ k } = δ j k \begin{align}
& \left\{\bar{q}_j, \bar{q}_k\right\}=\left\{\bar{p}_j, \bar{p}_k\right\}=0 \tag{2.8}\\
& \left\{\bar{q}_j, \bar{p}_k\right\}=\delta_{j k}\tag{2.9}
\end{align} { q ˉ j , q ˉ k } = { p ˉ j , p ˉ k } = 0 { q ˉ j , p ˉ k } = δ j k ( 2.8 ) ( 2.9 ) 因此，变换后的轨迹也满足Hamilton方程 (2.4) 和 (2.5)。
现在回到二维谐振子，以说明非正则变换会破坏上述结果。Hamilton量为
H = 1 2 m ( p x 2 + p y 2 ) + 1 2 m ω 2 ( x 2 + y 2 ) H=\frac{1}{2 m}\left(p_x^2+p_y^2\right)+\frac{1}{2} m \omega^2\left(x^2+y^2\right) H = 2 m 1 ( p x 2 + p y 2 ) + 2 1 m ω 2 ( x 2 + y 2 ) 考虑只旋转坐标而不旋转动量的变换。该变换为
x ˉ = x cos θ − y sin θ y ˉ = x sin θ + y cos θ p ˉ x = p x p ˉ y = p y \begin{aligned}
\bar{x} & =x \cos \theta-y \sin \theta \\
\bar{y} & =x \sin \theta+y \cos \theta \\
\bar{p}_x & =p_x \\
\bar{p}_y & =p_y
\end{aligned} x ˉ y ˉ p ˉ x p ˉ y = x cos θ − y sin θ = x sin θ + y cos θ = p x = p y 如前所见，这是一个非正则变换。为了看看会发生什么，考虑初始条件
x ( 0 ) = a p x ( 0 ) = b y ( 0 ) = p y ( 0 ) = 0 \begin{aligned}
x(0) & =a \\
p_x(0) & =b \\
y(0) & =p_y(0)=0
\end{aligned} x ( 0 ) p x ( 0 ) y ( 0 ) = a = b = p y ( 0 ) = 0 质点从 x x x x x x x x x
p ˙ x = − ∂ H ∂ x = − m ω 2 x x ˙ = ∂ H ∂ p x = p x m \begin{align}
\dot{p}_x & =-\frac{\partial H}{\partial x}=-m \omega^2 x \tag{2.10}\\
\dot{x} & =\frac{\partial H}{\partial p_x}=\frac{p_x}{m}\tag{2.11}
\end{align} p ˙ x x ˙ = − ∂ x ∂ H = − m ω 2 x = ∂ p x ∂ H = m p x ( 2.10 ) ( 2.11 ) 关于 y y y p y p_y p y x x x y y y
p ¨ x = − m ω 2 x ˙ = − ω 2 p x \ddot{p}_x=-m \omega^2 \dot{x}=-\omega^2 p_x p ¨ x = − m ω 2 x ˙ = − ω 2 p x 其通解为
p x ( t ) = A cos ω t + B sin ω t p_x(t)=A \cos \omega t+B \sin \omega t p x ( t ) = A cos ω t + B sin ω t 对 x x x
x ( t ) = C cos ω t + D sin ω t x(t)=C \cos \omega t+D \sin \omega t x ( t ) = C cos ω t + D sin ω t 代入初始条件，得到
p x ( 0 ) = A = b x ( 0 ) = C = a \begin{aligned}
p_x(0) & =A=b \\
x(0) & =C=a
\end{aligned} p x ( 0 ) x ( 0 ) = A = b = C = a 把这些结果代入运动方程 (2.10) 和 (2.11)，并解出 B B B D D D
p x ( t ) = b cos ω t − m ω a sin ω t x ( t ) = a cos ω t + b m ω sin ω t y ( t ) = p y ( t ) = 0 \begin{aligned}
p_x(t) & =b \cos \omega t-m \omega a \sin \omega t \\
x(t) & =a \cos \omega t+\frac{b}{m \omega} \sin \omega t \\
y(t) & =p_y(t)=0
\end{aligned} p x ( t ) x ( t ) y ( t ) = b cos ω t − mω a sin ω t = a cos ω t + mω b sin ω t = p y ( t ) = 0 现在假设初始条件为 x ( 0 ) = 0 x(0)=0 x ( 0 ) = 0 y ( 0 ) = a y(0)=a y ( 0 ) = a p x ( 0 ) = b p_x(0)=b p x ( 0 ) = b p y ( 0 ) = 0 p_y(0)=0 p y ( 0 ) = 0 π 2 \frac{\pi}{2} 2 π y y y x x x x x x y y y π 2 \frac{\pi}{2} 2 π
p x ( t ) = b cos ω t p y ( t ) = − m ω a sin ω t x ( t ) = b m ω sin ω t y ( t ) = a cos ω t \begin{aligned}
p_x(t) & =b \cos \omega t \\
p_y(t) & =-m \omega a \sin \omega t \\
x(t) & =\frac{b}{m \omega} \sin \omega t \\
y(t) & =a \cos \omega t
\end{aligned} p x ( t ) p y ( t ) x ( t ) y ( t ) = b cos ω t = − mω a sin ω t = mω b sin ω t = a cos ω t 例如考察 t = π 2 ω t=\frac{\pi}{2 \omega} t = 2 ω π cos ω t = 0 \cos \omega t=0 cos ω t = 0 sin ω t = 1 \sin \omega t=1 sin ω t = 1 x x x ( x , y ) = ( b m ω , 0 ) (x, y)=\left(\frac{b}{m \omega}, 0\right) ( x , y ) = ( mω b , 0 ) y y y
另一种可能更容易理解的方式是：由于第一个质点始终只沿 x x x y y y y y y p x ≠ 0 p_x \neq 0 p x = 0
一般而言，如果变换不是正则变换，那么 (2.6) 和 (2.7) 中的Poisson括号就不满足条件 (2.8) 和 (2.9)，结果是Hamilton方程在 ( q ˉ , p ˉ ) (\bar{q}, \bar{p}) ( q ˉ , p ˉ )
■ ~\tag*{$\blacksquare$} ■ 习题 2.8.6 证明 ∂ S c l / ∂ x f = p ( t f ) \partial S_{\mathrm{cl}} / \partial x_f=p\left(t_f\right) ∂ S cl / ∂ x f = p ( t f )
解答: 情形如下图所示：
现在两条轨迹经历的时间相同，但在修正后的轨迹中，粒子多运动了一段距离 Δ x \Delta x Δ x L Δ t \mathscr{L}\Delta t L Δ t η ( t ) > 0 \eta(t)>0 η ( t ) > 0 x ( t ) x(t) x ( t ) x c l ( t ) x_{\mathrm{cl}}(t) x cl ( t )
δ S c l = ∂ L ∂ x ˙ η ( t ) ∣ t f \delta S_{\mathrm{cl}}=\left.\frac{\partial \mathscr{L}}{\partial \dot{x}} \eta(t)\right|_{t_f} δ S cl = ∂ x ˙ ∂ L η ( t ) t f 在 t = t f t=t_{f} t = t f η ( t f ) = Δ x \eta(t_{f})=\Delta x η ( t f ) = Δ x
δ S c l = ∂ L ∂ x ˙ ∣ t f Δ x ∂ S c l ∂ x f = ∂ L ∂ x ˙ ∣ t f = p ( t f ) \begin{aligned}
\delta S_{\mathrm{cl}}&=\left.\frac{\partial \mathscr{L}}{\partial \dot{x}}\right|_{t_{f}}\Delta x\\
\frac{\partial S_{\mathrm{cl}}}{\partial x_{f}}&=\left.\frac{\partial \mathscr{L}}{\partial \dot{x}}\right|_{t_{f}}=p(t_{f})
\end{aligned} δ S cl ∂ x f ∂ S cl = ∂ x ˙ ∂ L t f Δ x = ∂ x ˙ ∂ L t f = p ( t f ) ■ ~\tag*{$\blacksquare$} ■ 习题 2.8.7 考虑谐振子，其通解为
x ( t ) = A cos ω t + B sin ω t . x(t)=A \cos \omega t+B \sin \omega t . x ( t ) = A cos ω t + B sin ω t . 将能量用 A A A B B B A A A B B B x ( 0 ) = x 1 x(0)=x_1 x ( 0 ) = x 1 x ( T ) = x 2 x(T)=x_2 x ( T ) = x 2 x 1 , x 2 , T x_1,x_2,T x 1 , x 2 , T x 1 x_1 x 1 x 2 x_2 x 2
S c l ( x 1 , x 2 , T ) = m ω 2 sin ω T [ ( x 1 2 + x 2 2 ) cos ω T − 2 x 1 x 2 ] S_{\mathrm{cl}}\left(x_1, x_2, T\right)=\frac{m \omega}{2 \sin \omega T}\left[\left(x_1^2+x_2^2\right) \cos \omega T-2 x_1 x_2\right] S cl ( x 1 , x 2 , T ) = 2 sin ω T mω [ ( x 1 2 + x 2 2 ) cos ω T − 2 x 1 x 2 ] 验证 ∂ S c l / ∂ T = − E \partial S_{\mathrm{cl}} / \partial T=-E ∂ S cl / ∂ T = − E
解答: 对于一维谐振子，有
∂ S c l ∂ t f = − H ( t f ) \frac{\partial S_{\mathrm{cl}}}{\partial t_f}=-H\left(t_f\right) ∂ t f ∂ S cl = − H ( t f ) 位置的通解为
x ( t ) = A cos ω t + B sin ω t x ˙ ( t ) = − A ω sin ω t + B ω cos ω t \begin{aligned}
& x(t)=A \cos \omega t+B \sin \omega t \\
& \dot{x}(t)=-A \omega \sin \omega t+B \omega \cos \omega t
\end{aligned} x ( t ) = A cos ω t + B sin ω t x ˙ ( t ) = − A ω sin ω t + B ω cos ω t 总能量为
E = 1 2 m x ˙ 2 + 1 2 m ω 2 x 2 = m 2 ( ( − A ω sin ω t + B ω cos ω t ) 2 + ω 2 ( A cos ω t + B sin ω t ) 2 ) = m ω 2 2 ( A 2 + B 2 ) (2.12) \begin{aligned}
E & =\frac{1}{2} m \dot{x}^2+\frac{1}{2} m \omega^2 x^2 \\
& =\frac{m}{2}\left((-A \omega \sin \omega t+B \omega \cos \omega t)^2+\omega^2(A \cos \omega t+B \sin \omega t)^2\right) \\
& =\frac{m \omega^2}{2}\left(A^2+B^2\right)\tag{2.12}
\end{aligned} E = 2 1 m x ˙ 2 + 2 1 m ω 2 x 2 = 2 m ( ( − A ω sin ω t + B ω cos ω t ) 2 + ω 2 ( A cos ω t + B sin ω t ) 2 ) = 2 m ω 2 ( A 2 + B 2 ) ( 2.12 ) 这里我们只是将第二行展开，约去相消项，并使用了 cos 2 x + sin 2 x = 1 \cos ^2 x+\sin ^2 x=1 cos 2 x + sin 2 x = 1
为了求作用量，需要Lagrange量：
L = T − V = 1 2 m x ˙ 2 − 1 2 m ω 2 x 2 = m 2 ( ( − A ω sin ω t + B ω cos ω t ) 2 − ω 2 ( A cos ω t + B sin ω t ) 2 ) = m ω 2 2 [ A 2 ( sin 2 ω t − cos 2 ω t ) + B 2 ( cos 2 ω t − sin 2 ω t ) − 4 A B sin ω t cos ω t ] = m ω 2 2 ( ( B 2 − A 2 ) cos 2 ω t − 2 A B sin 2 ω t ) \begin{aligned}
L & =T-V \\
& =\frac{1}{2} m \dot{x}^2-\frac{1}{2} m \omega^2 x^2 \\
& =\frac{m}{2}\left((-A \omega \sin \omega t+B \omega \cos \omega t)^2-\omega^2(A \cos \omega t+B \sin \omega t)^2\right) \\
& =\frac{m \omega^2}{2}\left[A^2\left(\sin ^2 \omega t-\cos ^2 \omega t\right)+B^2\left(\cos ^2 \omega t-\sin ^2 \omega t\right)-4 A B \sin \omega t \cos \omega t\right] \\
& =\frac{m \omega^2}{2}\left(\left(B^2-A^2\right) \cos 2 \omega t-2 A B \sin 2 \omega t\right)
\end{aligned} L = T − V = 2 1 m x ˙ 2 − 2 1 m ω 2 x 2 = 2 m ( ( − A ω sin ω t + B ω cos ω t ) 2 − ω 2 ( A cos ω t + B sin ω t ) 2 ) = 2 m ω 2 [ A 2 ( sin 2 ω t − cos 2 ω t ) + B 2 ( cos 2 ω t − sin 2 ω t ) − 4 A B sin ω t cos ω t ] = 2 m ω 2 ( ( B 2 − A 2 ) cos 2 ω t − 2 A B sin 2 ω t ) 从 t = 0 t=0 t = 0 t = T t=T t = T
S = ∫ 0 T L d t = m ω 4 [ ( B 2 − A 2 ) sin 2 ω t + 2 A B cos 2 ω t ] 0 T = m ω 4 [ ( B 2 − A 2 ) sin 2 ω T + 2 A B ( cos 2 ω T − 1 ) ] = m ω 2 [ ( B 2 − A 2 ) sin ω T cos ω T + A B ( cos 2 ω T − sin 2 ω T − 1 ) ] = m ω 2 [ ( B 2 − A 2 ) sin ω T cos ω T − 2 A B sin 2 ω T ] (2.13) \begin{aligned}
S & =\int_0^T L d t \\
& =\frac{m \omega}{4}\left[\left(B^2-A^2\right) \sin 2 \omega t+2 A B \cos 2 \omega t\right]_0^T \\
& =\frac{m \omega}{4}\left[\left(B^2-A^2\right) \sin 2 \omega T+2 A B(\cos 2 \omega T-1)\right] \\
& =\frac{m \omega}{2}\left[\left(B^2-A^2\right) \sin \omega T \cos \omega T+A B\left(\cos ^2 \omega T-\sin ^2 \omega T-1\right)\right] \\
& =\frac{m \omega}{2}\left[\left(B^2-A^2\right) \sin \omega T \cos \omega T-2 A B \sin ^2 \omega T\right]\tag{2.13}
\end{aligned} S = ∫ 0 T L d t = 4 mω [ ( B 2 − A 2 ) sin 2 ω t + 2 A B cos 2 ω t ] 0 T = 4 mω [ ( B 2 − A 2 ) sin 2 ω T + 2 A B ( cos 2 ω T − 1 ) ] = 2 mω [ ( B 2 − A 2 ) sin ω T cos ω T + A B ( cos 2 ω T − sin 2 ω T − 1 ) ] = 2 mω [ ( B 2 − A 2 ) sin ω T cos ω T − 2 A B sin 2 ω T ] ( 2.13 ) 为了继续计算，需要确定 A A A B B B t = 0 t=0 t = 0 t = T t=T t = T x ( 0 ) = x 1 x(0)=x_1 x ( 0 ) = x 1 x ( T ) = x 2 x(T)=x_2 x ( T ) = x 2
A = x 1 x 1 cos ω T + B sin ω T = x 2 B = x 2 − x 1 cos ω T sin ω T \begin{aligned}
A & =x_1 \\
x_1 \cos \omega T+B \sin \omega T & =x_2 \\
B & =\frac{x_2-x_1 \cos \omega T}{\sin \omega T}
\end{aligned} A x 1 cos ω T + B sin ω T B = x 1 = x 2 = sin ω T x 2 − x 1 cos ω T 将这些结果代入 (2.12)，得到能量
E = m ω 2 2 ( x 1 2 + ( x 2 − x 1 cos ω T sin ω T ) 2 ) = m ω 2 2 sin 2 ω T ( x 1 2 + x 2 2 − 2 x 1 x 2 cos ω T ) \begin{aligned}
E & =\frac{m \omega^2}{2}\left(x_1^2+\left(\frac{x_2-x_1 \cos \omega T}{\sin \omega T}\right)^2\right) \\
& =\frac{m \omega^2}{2 \sin ^2 \omega T}\left(x_1^2+x_2^2-2 x_1 x_2 \cos \omega T\right)
\end{aligned} E = 2 m ω 2 ( x 1 2 + ( sin ω T x 2 − x 1 cos ω T ) 2 ) = 2 sin 2 ω T m ω 2 ( x 1 2 + x 2 2 − 2 x 1 x 2 cos ω T ) 将 A A A B B B
S = m ω 2 sin ω T [ ( x 2 − x 1 cos ω T ) 2 cos ω T − x 1 sin 2 ω T cos ω T − 2 x 1 sin 2 ω T ( x 2 − x 1 cos ω T ) ] = m ω 2 sin ω T [ ( x 2 2 − 2 x 1 x 2 cos ω T + x 1 2 cos 2 ω T ) cos ω T − x 1 2 sin 2 ω T cos ω T − 2 x 1 x 2 sin 2 ω T + 2 x 1 sin 2 ω T cos ω T ] = m ω 2 sin ω T [ ( x 1 2 + x 2 2 ) cos ω T − 2 x 1 x 2 ] \begin{aligned}
S & =\frac{m \omega}{2 \sin\omega T}\left[\left(x_2-x_1 \cos\omega T\right)^2 \cos\omega T-x_1 \sin^2\omega T \cos\omega T-2 x_1 \sin^2\omega T\left(x_2-x_1 \cos\omega T\right)\right] \\
& =\frac{m \omega}{2 \sin\omega T}[\left(x_2^2-2 x_1 x_2 \cos\omega T+x_1^2 \cos^2\omega T\right) \cos\omega T-x_1^2 \sin^2\omega T \cos\omega T\\
&~~~~~~~~~~~~~~~~~~~~-2 x_1 x_2 \sin^2\omega T+2 x_1 \sin^2\omega T \cos\omega T] \\
& =\frac{m \omega}{2 \sin \omega T}\left[\left(x_1^2+x_2^2\right) \cos \omega T-2 x_1 x_2\right]
\end{aligned} S = 2 sin ω T mω [ ( x 2 − x 1 cos ω T ) 2 cos ω T − x 1 sin 2 ω T cos ω T − 2 x 1 sin 2 ω T ( x 2 − x 1 cos ω T ) ] = 2 sin ω T mω [ ( x 2 2 − 2 x 1 x 2 cos ω T + x 1 2 cos 2 ω T ) cos ω T − x 1 2 sin 2 ω T cos ω T − 2 x 1 x 2 sin 2 ω T + 2 x 1 sin 2 ω T cos ω T ] = 2 sin ω T mω [ ( x 1 2 + x 2 2 ) cos ω T − 2 x 1 x 2 ] 对其求导，得到
∂ S ∂ T = m ω 2 sin 2 ω T [ − ω ( x 1 2 + x 2 2 ) sin 2 ω T − ( ( x 1 2 + x 2 2 ) cos ω T − 2 x 1 x 2 ) ω cos ω T ] = m ω 2 2 sin 2 ω T [ − ( x 1 2 + x 2 2 ) + 2 x 1 x 2 cos ω T ] = − m ω 2 2 sin 2 ω T ( x 1 2 + x 2 2 − 2 x 1 x 2 cos ω T ) = − E \begin{aligned}
\frac{\partial S}{\partial T} & =\frac{m \omega}{2 \sin^2\omega T}\left[-\omega\left(x_1^2+x_2^2\right) \sin^2\omega T-\left(\left(x_1^2+x_2^2\right) \cos\omega T-2 x_1 x_2\right) \omega \cos\omega T\right] \\
& =\frac{m \omega^2}{2 \sin^2\omega T}\left[-\left(x_1^2+x_2^2\right)+2 x_1 x_2 \cos\omega T\right] \\
& =-\frac{m \omega^2}{2 \sin ^2 \omega T}\left(x_1^2+x_2^2-2 x_1 x_2 \cos \omega T\right) \\
& =-E
\end{aligned} ∂ T ∂ S = 2 sin 2 ω T mω [ − ω ( x 1 2 + x 2 2 ) sin 2 ω T − ( ( x 1 2 + x 2 2 ) cos ω T − 2 x 1 x 2 ) ω cos ω T ] = 2 sin 2 ω T m ω 2 [ − ( x 1 2 + x 2 2 ) + 2 x 1 x 2 cos ω T ] = − 2 sin 2 ω T m ω 2 ( x 1 2 + x 2 2 − 2 x 1 x 2 cos ω T ) = − E 因此，对于谐振子，所要求的结果得到了验证。
■ ~\tag*{$\blacksquare$} ■ 
