两体中心势

设两质点质量 \(m_1, m_2\), 位置矢量为 \(\vec{r}_1, \vec{r}_2\). 两者相互作用力方向相反,在两体的连线上, 大小相等且依赖于两质点间距离: \(r = \sqrt{\vec{r}\cdot \vec{r}},\; \vec{r} = \vec{r}_2 - \vec{r}_1\). \[\vec{f}_{21} = f(r) \hat{r} = -\vec{f}_{12}.\] 易证这种作用力是保守力, 其相互作用力可以表示成势函数梯度: \[\vec{f}_{21} = -\frac{\partial U(r)}{\partial \vec{r}_2}, \;\vec{f}_{12} = -\frac{\partial U(r)}{\partial \vec{r}_1}.\] 体系的拉格朗日量: \[L = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 - U(r).\] 取质心坐标 \(\vec{R}\) 和相对坐标 \(\vec{r}\) 为广义坐标: \[\vec{R} = \frac{1}{M}\left( m_1 \vec{r}_1 + m_2\vec{r}_2\right), \;\vec{r} = \vec{r}_2 - \vec{r}_1.\] 其中 \(M = m_1 + m_2\). 可算得 \(\vec{r}_1, \vec{r}_2\) 与 \(\vec{R}, \vec{r}\) 关系为:

\begin{cases} \vec{r}_1 = \vec{R} - \frac{m_2}{M}\vec{r} = \vec{R} - \frac{\mu}{m_1}\vec{r},\\ \vec{r}_2 = \vec{R} + \frac{m_1}{M}\vec{r} = \vec{R} + \frac{\mu}{m_2}\vec{r}. \end{cases}

其中 \(\mu\) 满足 \[\frac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}.\] 故可得: 当 \(m_1 \gg \ m_2\) 时, 有近似: \(\vec{r}_1\approx \vec{R}, \vec{r}_2\approx \vec{R}+\vec{r}.\)

新拉格朗日量为: \[L = \frac{1}{2}MV^2 + \frac{1}{2}\mu v^2 - U(r).\] 显然 \(\vec{R}\) 是循环坐标, 故 \(\mathrm{d}_t\vec{V} = 0\Rightarrow \vec{V} = \mathbf{0}\). 因此拉格朗日量退化为: \[L = \frac{1}{2}\mu v^2 - U(r) \stackrel{(r,\theta,\phi)}{=} \frac{1}{2}\mu \left( \dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta \;\dot{\phi}^2 \right) - U(r).\] 取初始时刻 \(\theta_0 = \frac{\pi}{2}, \dot{\theta}_0 = 0\). \[\frac{\mathrm{d}}{\mathrm{d}t}\left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 \Rightarrow \frac{\mathrm{d}}{\mathrm{d}t} \left( r^2\dot{\theta} \right)- r^2\sin\theta\cos\theta\;\dot{\phi}^2 = 0.\] 整理可得: \[2r\dot{r}\dot{\theta} + r^2\ddot{\theta} - r^2\sin\theta\cos\theta\;\dot{\phi}^2.\] 代入初始条件可得: \(\ddot{\theta}_0 = 0\), 进一步可求得:

\begin{cases}\theta = \frac{\pi}{2},\\ \dot{\theta}=2,\\ \ddot{\theta} = 0.\end{cases}

代入拉格朗日方程: \[L=\frac{1}{2}\mu \left( \dot{r}^2+r^2\dot{\phi}^2 \right)-U(r).\]

从上一节得到拉格朗日方程出发, \[L=\frac{1}{2}\mu \left( \dot{r}^2+r^2\dot{\phi}^2 \right)-U(r).\] 易证\(\phi\)是可遗坐标. 可得: \[\mu \frac{\mathrm{d}}{\mathrm{d}t}\left( r^2\vec{\phi} \right)=0 \Rightarrow\mu r^2\dot{\phi} = l_0 = \text{const}.\] 在 \(\mathrm{d}t\) 时间内 \(\vec{r}\) 矢量扫过面积为 \[\mathrm{d}A = \frac{1}{2}r\mathrm{d}r = \frac{1}{2}r^2\mathrm{d}\phi = \frac{1}{2}r^2\dot{\phi}\mathrm{d}t.\] 可得: \[\frac{\mathrm{d}A}{\mathrm{d}t}=\frac{r^2\dot{\phi}}{2}=\frac{l_0}{2\mu}.\]

开普勒第二定律

\(\vec{r}\) 矢量在单位时间扫过相同面积.

广义能量函数有: \[H=\frac{1}{2}\mu \left( \dot{r}^2 + r^2\dot{\phi}^2 \right) + U(r).\] 由广义功能定理可知: \[\frac{\mathrm{d}H}{\mathrm{d}t}=-\frac{\partial L}{\partial t}=0.\Rightarrow H=E_0.\] 由前述 \(\mu r^2\dot{\phi} = l_0\), 代入哈密顿量有: \[H=\frac{1}{2}\mu \dot{r}^2 + \left[ \frac{l_0^2}{2\mu r^2} + U(r) \right] = \frac{1}{2}\mu \dot{r}^2 + U_{\text{eff}}(r).\] 且 \(H = E_0\), 得到

\[\dot{r} = \pm \sqrt{\frac{2}{\mu}\left[ E_0 - U_{\text{eff}}(r) \right]}.\] 其中 \(\pm\) 在当处在\(r\)增加的轨道部分时取正, 反之取负.

考虑到径向速度 \(\dot{r}\in \mathbb{R}\), \(E_0\ge U_{\text{eff}}(r).\), 可得: \[t = \pm \sqrt{\frac{\mu}{2}} \int\left[ E_0-U_{\text{eff}}(r) \right]^{-1/2}\mathrm{d}r + C_1\] 也可以用于求周期. 给定 \(U(r)\), 可得 \(t = t(r)\Rightarrow r = r(t)\).

由 \(\mu r^2\dot{\phi}^2 = l_0\) 可得: \[\phi(t) = \frac{l_0}{\mu}\int \frac{\mathrm{d}}{r^2(t)} + C_2.\]

另外已知 \(\dot{r} 和 \dot{\phi}\), 可得 \[\frac{\mathrm{d}\phi}{\mathrm{d}r} = \frac{\dot{\phi}}{\dot{r}} = \pm \sqrt{\frac{l_0^2}{2\mu}}\frac{1}{r^2 \sqrt{E_0-U_{\text{eff}}}}.\] 得到 \(\phi(r)\):

\[\phi(r) = \pm\sqrt{\frac{l_0^2}{2\mu}} \int\frac{\mathrm{d}r}{r^2 \sqrt{E_0-U_{\text{eff}}}} + C_3.\] 有逆函数 \(r = r(\phi)\), 进一步得到 \(t(\phi)\): \[t = \frac{\mu}{l_0}\int r^2(\phi)\mathrm{d}\phi + C_4.\]

有效势能 \(U_{\text{eff}}(r) = U(r) + \frac{l_0^2}{2\mu r^2}\), 机械能 \(E_0 = \frac{\mu \dot{r}}{2} + U_{\text{eff}}(r) \ge U_{\text{eff}}(r)\).

圆轨道的稳定性

• 存在性: \[\frac{\mathrm{d}U_{\text{eff}}}{\mathrm{d}r}\Big{|}_{r_m} = 0.\]
• 稳定性: \[\frac{\mathrm{d}^2 U_{\text{eff}}}{\mathrm{d}r^2}\Big{|}_{r_m}>0.\]

\[U(r) = \pm \frac{k}{r},\;k>0.\] 代入 \(\phi(r)\) 方程, 可得: \[\phi = \pm \arccos{\left(\frac{l_0^2}{\mu k\varepsilon r} \pm \frac{1}{\varepsilon}\right)} + C_3.\] 其中 \[\varepsilon = \sqrt{1 + \frac{2E_0 l_0^2}{\mu k^2}}.\] 可得逆函数 \(r(\phi)\): \[r = \left( \frac{l_0^2}{\mu k} \right)\frac{1}{\varepsilon\cos(\phi - C_3)\mp 1}.\] 取 \(C_3 = \min\phi\), 当 \(\phi = \min\phi\) 时有 \(r = \min r\). 易证:

开普勒第一定律

\(r\) 轨道为圆锥曲线.

由开普勒第二定律知 \[\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{l_0}{2\mu},\] 又 ∵ \(r\) 轨道闭合时为椭圆, 可得周期: \[\tau = \frac{\pi ab}{\mathrm{d}A/\mathrm{d}t} = 2\pi \sqrt{\frac{\mu}{k}}a^{3/2}.\] 其中 \(\mu = \frac{Mm}{Mm}, k = GMm\), 又有 \(M\gg m\), 有近似: \[\frac{\tau^2}{a^3} = \frac{4\pi^2}{G(M+m)}\approx \frac{4\pi^2}{GM}.\]

开普勒第三定律

公转周期的平方及其椭圆轨道半长轴的立方成正比.

何时轨道封闭,若有进动,要几圈才封闭

Bertrand 定理

仅对 \[U(r) = -\frac{\kappa}{r},\;U(r) = \kappa r^2,\;(\kappa>0)\] 两种有心势, 其所有有限轨道闭合.

以下给出证明. 轨道闭合等价于返回 \(\min r\) 时, \(\phi\)的增量\(\Delta\phi\)与\(2\pi\)之比应为有理数.

将 \(r = \frac{1}{u}\) 代入 \(\mathrm{d}_r\phi\) 方程: \[\Delta\phi = 2 \int_{\min u}^{\max u} \frac{l_0 \mathrm{d}u} {\sqrt{2\mu \left[ E - U_{\text{eff}}\left( \frac{1}{u} \right) \right]}}\] 且 \(n\Delta\phi = m 2\pi\), \(n, m \in \mathbb{Z}\). 取 \(W(u) = U_{\text{eff}}\left( \frac{1}{u} \right)\). \(U_{\text{eff}}\) 极小点为 \(r_0\), \(u_0 = \frac{1}{r_0}\). 对 \(E - U = E - W\) 进行 Taylor 展开: \[E - W(u) = E_0 +\Delta E - W(u_0) + W^{\prime}(u_0)(u-u_0) - W^{\prime\prime}(u_0)(u-u_0)^2.\] 显然 \(W^{\prime}(u_0) = 0, W^{\prime\prime}(u_0)>0\), 且 \(\dot{r}_0 = 0\Rightarrow E_0 = W(u_0)\). 可得: \[E - W(u) = \Delta E - \frac{1}{2}W^{\prime\prime}(u-u_0)^2, \;W^{\prime\prime}(u_0) = r_0^4 U^{\prime\prime}(r_0) > 0.\] 得到: \[\Delta\phi = \frac{2l_0}{\sqrt{\mu W^{\prime\prime}(u_0)}}\int_{u_{\min}}^{u_{\max}} \frac{l \mathrm{d}u}{\sqrt{\frac{2\Delta E}{W^{\prime\prime}(u_0)} - (u-u_0^2)}}.\] 最终算得: \[\Delta\phi = \frac{2\pi}{\sqrt{\beta + 2}}, \;U = \kappa r^{\beta}\Rightarrow\beta = -1, 2.\blacksquare\]

Bertrand 定理证明:

• 对于平方反比引力势 \(U = -|\kappa|/r\), \(\Delta\phi = 2\pi\), \(r\) 的一次来回就能闭合.
• 对于虎克势 \(U = \kappa r^2\), \(\Delta\phi = \pi\), \(r\) 的两次来回才能闭合.

假设散射过程粒子内部状态不变, 可得: 散射前后粒子总动量、总能量守恒. 即: \[\frac{\mathrm{d}\vec{V}}{\mathrm{d}t} = \ddot{\vec{R}} = 0.\] 质心系是惯性系. 有动量守恒:

\begin{equation*} m_1 \vec{v}_{1i}^{\prime} + m_2 \vec{v}_{2i}^{\prime} = m_1 \vec{v}_{1f}^{\prime} + m_2 \vec{v}_{2f}^{\prime} = 0 \Rightarrow \begin{cases} m_2\vec{v}_{2i}^{\prime} = -m_1\vec{v}^{\prime}_{1i},\\ m_2\vec{v}_{2f}^{\prime} = -m_1\vec{v}^{\prime}_{1f} \end{cases} \end{equation*}

其中 \(i, f\) 分别表示散射前后初态与末态. 联立能量守恒方程:

\begin{equation*} \frac{m_1}{2}{\vec{v}_{1i}^{\prime}}^2 + \frac{m_2}{2}{\vec{v}_{2i}^{\prime}}^2 = \frac{m_1}{2}{\vec{v}_{1f}^{\prime}}^2 + \frac{m_2}{2}{\vec{v}_{2f}^{\prime}}^2 \Rightarrow \begin{cases} v^{\prime}_{1i} = v^{\prime}_{1f},\\ v^{\prime}_{2i} = v^{\prime}_{2f} \end{cases} \end{equation*}

各自速率大小不变. 在实验系坐标系中有: \[\vec{v}_1 = \vec{V} + \vec{v}^{\prime}_1, \;\vec{v}_2 = \vec{V} + \vec{v}^{\prime}_2\] 相对速度有: \[\vec{v} = \vec{v}_1 - \vec{v}_2 = \vec{v}_1^{\prime}-\vec{v}_2^{\prime}\] 下左图即为质心坐标系中的散射.

上右图为初始时刻靶粒子静止的情况. 此时有: \(\vec{v}_{2i} = \vec{0},\;\vec{v}_{2i}^{\prime} = -\vec{V}\). 可得 \[m_1\vec{v}^{\prime}_{1i} = -m_2\vec{v}_{2i}^{\prime} = m_2 \vec{V}, \;\vec{v}_{1i} = \vec{v}^{\prime}_{1i} + \vec{V} \Rightarrow \vec{v}_{1i}\parallel \vec{V}.\]

• \(\Theta_1\) 为实验系入射粒子散射前后速度偏转角 (散射角),
• \(\Theta_2\) 为靶粒子速度偏转角 (反冲角).

由图可知:

\begin{cases} \tan\Theta_1 = \frac{v_{1f}^{\prime}\sin\theta}{V + v_{1f}^{\prime}\cos\theta} = \frac{\sin\theta}{(m_1/m_2)+\cos\theta},\\ \Theta_2 = \frac{\pi - \theta}{2}. \end{cases}

取靶粒子固定, 即 \(m_2 \gg m_1\) 可得 \(\Theta_1 = \theta, 2\Theta_2 + \theta =\pi\).

下图为弹性散射的几何参数. 其中 \(b\) 是

碰撞参数

入射线到靶粒子距离 \(b\).

能解得: \[\phi_1 = \frac{l^2}{2\mu}\int_{r_{\min}}^{\infty} \frac{2\mathrm{d}r}{r^2 \sqrt{E - U_{\text{eff}}}} = 2\varphi.\] 由此可得质心系散射角 \[\theta = \pi - \phi_1 = \pi - 2\varphi.\]

\[\mathrm{d}\sigma = \frac{b}{\sin\theta} \left| \frac{\partial(b, \phi_b)}{\partial(\theta, \phi)} \right| \mathrm{d}\Omega = \sigma(\theta, \phi)\mathrm{d}\Omega.\] 解得 \[\sigma(\theta,\phi) = \frac{b}{\sin\theta} \left| \frac{\partial(b, \phi_b)}{\partial(\theta, \phi)} \right|.\] 从对称性可知\(b\)不是依赖于\(\phi\)的函数. 可得: \[\sigma(\theta, \phi) = \frac{b(\theta)}{\sin\theta} \left| \frac{\mathrm{d}b(\theta)}{\mathrm{d}\theta} \right|.\]

• 由角动量守恒可得: \[\mu r^2\vec{\phi} = l = \mu v b.\]
• 由能量守恒: \[\dot{r}=\beta\left[\frac{2}{\mu}\left[E - U_{\text{eff}(r)}\right]\right].\] 其中 \(E = \frac{\mu}{2}v_0^2, l=\mu bv_0\), 可得 \(l^2 = 2\mu b^2 E\).

已知有心力势是库仑势 \(U = \frac{\alpha}{r}\). 此时 \(r_{\min}\) 满足 \(\dot{r} = 0,\;E=U_{\text{eff}}\), 可得: \[E = \frac{\alpha}{r_{\min}} + \frac{l^2}{2\mu r_{\min}^2} = \frac{\alpha}{r_{\min}} + \frac{b^2 E}{r_{\min}^2}.\] 可得 \(r_{\min}\) 满足: \[\frac{1}{r_{\min}} = \frac{1}{b^2} \left[\sqrt{\left(\frac{\alpha}{2E}\right)^2+b^2}-\frac{\alpha}{2E}\right].\] \(\varphi\) 即有:

\begin{aligned} \varphi &= \sqrt{\frac{l^2}{2\mu}}\int_{r_{\min}}^{\infty} \frac{\mathrm{d}r}{r^2 \sqrt{E - U_{\text{eff}}}}\\ &= b \sqrt{E}\int_0^{r_{\min}^{-1}} \frac{\mathrm{d}u}{\sqrt{E-\left( \alpha u + \frac{l^2}{2\mu^2}u^2 \right)}}\\ &=\arccos \left( \frac{\alpha/E}{\sqrt{4b^2 + (\alpha/E)^2}} \right) \end{aligned}

故有: \[\tan\varphi = \frac{2bE}{\alpha}.\] 又因为 \(\theta = \pi - 2\varphi\), 有: \[\tan \frac{\theta}{2} = \frac{\alpha}{2bE} \Rightarrow b=\frac{\alpha}{2E}\cot \frac{\theta}{2}.\] 得到库仑势下微分散射截面: \[\sigma(\theta, \phi) = \frac{b(\theta)}{\sin\theta} \left| \frac{\mathrm{d}b(\theta)}{\mathrm{d}\theta} \right| = \left( \frac{\alpha}{4E} \right)^2 \frac{1}{\sin^4 \frac{\theta}{2}}.\] 上式即 Rutherford 公式.