电动力学 第6周作业

Chasse_neige

1. 书6.12, 6.15, 6.16

6.12 电偶极子 \(p_0\) 以速度 \(v\) 作匀速运动，求它产生的电磁势和场 \(\varphi, A; E, B\)。

取Lorentz Boost的方向为x轴，记 \(p_{0}\) 所在位置为原点，记当前时刻为时间零点。

电磁势：

在电偶极子系中，电偶极子产生的电磁势为： $$ \begin{aligned} \begin{pmatrix} A_{x} \\ A_{y} \\ A_{z} \\ \frac{\phi}{c} \end{pmatrix} = \begin{pmatrix} \gamma & 0 & 0 & \gamma \beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \gamma \beta & 0 & 0 & \gamma \end{pmatrix} \begin{pmatrix} A_{x}' \\ A_{y}' \\ A_{z}' \\ \frac{\phi'}{c} \end{pmatrix} = \begin{pmatrix} \gamma & 0 & 0 & \gamma \beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \gamma \beta & 0 & 0 & \gamma \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 0 \\ \frac{1}{4 \pi \epsilon_{0} c} \frac{\vec{r'} \cdot \vec{p_{0}}}{r'^{3}} \end{pmatrix} = \begin{pmatrix} \frac{\gamma \beta}{4 \pi \epsilon_{0} c} \frac{\vec{r'} \cdot \vec{p_{0}}}{r'^{3}} \\ 0 \\ 0 \\ \frac{\gamma}{4 \pi \epsilon_{0} c} \frac{\vec{r'} \cdot \vec{p_{0}}}{r'^{3}} \end{pmatrix} \end{aligned} $$

所以\(\phi = \frac{\gamma}{4 \pi \epsilon_{0}} \frac{\vec{r'} \cdot \vec{p_{0}}}{r'^{3}}\)， \(\vec{A} = \begin{pmatrix} \frac{\gamma \beta}{4 \pi \epsilon_{0} c} \frac{\vec{r'} \cdot \vec{p_{0}}}{r'^{3}}, 0, 0 \end{pmatrix}^{\top}\)

其中\(\vec{r'} = \begin{pmatrix} \gamma (x - \beta c t), y, z \end{pmatrix}^{\top}\)

在电偶极子系中，电偶极子产生的电场为： $$ \vec{E} = \frac{1}{4 \pi \epsilon_{0}} \frac{3(\vec{p_{0}} \cdot \vec{r'}) \vec{r'} - r'^{2} \vec{p_{0}}}{r'^{5}} $$ 换至地面系中，得到： $$ E_{x} = E_{x}' = \frac{1}{4 \pi \epsilon_{0}} \frac{3 (\vec{p_{0}} \cdot \vec{r'}) x' - r'^{2} (\vec{p_{0}} \cdot \hat{x'})}{r'^{5}} $$

\[ E_{y} = \gamma (E'_{y} + \beta c B'_{z}) = \gamma E'_{y} = \frac{\gamma}{4 \pi \epsilon_{0}} \frac{3(\vec{p_{0}} \cdot \vec{r'}) y' - r'^{2} (\vec{p_{0}} \cdot \hat{y'})}{r'^{5}} \]

\[ E_{z} = \gamma (E'_{z} - \beta c B'_{y}) = \gamma E'_{z} = \frac{\gamma}{4 \pi \epsilon_{0}} \frac{3(\vec{p_{0}} \cdot \vec{r'}) z' - r'^{2} (\vec{p_{0}} \cdot \hat{z'})}{r'^{5}} \]

\[ B_{x} = B'_{x} = 0 \]

\[ B_{y} = \gamma (B_{y'} - \beta \frac{E'_{z}}{c}) = - \frac{\gamma}{\beta c} E'_{z} = - \frac{\gamma}{4 \pi \epsilon_{0} \beta c} \frac{3(\vec{p_{0}} \cdot \vec{r'}) z' - r'^{2} (\vec{p_{0}} \cdot \hat{z'})}{r'^{5}} \]

\[ B_{z} = \gamma (B_{z'} + \beta \frac{E'_{y}}{c}) = \frac{\gamma}{\beta c} E'_{y} = \frac{\gamma}{4 \pi \epsilon_{0} \beta c} \frac{3(\vec{p_{0}} \cdot \vec{r'}) y' - r'^{2} (\vec{p_{0}} \cdot \hat{y'})}{r'^{5}} \]

其中\(\vec{r'} = \begin{pmatrix} \gamma (x - \beta c t), y, z \end{pmatrix}^{\top}\)

6.15 有一沿 \(z\) 轴方向螺旋进动的静磁场 \(\vec{B} = B_0 (\cos k_m z \hat{e}_x + \sin k_m z \hat{e}_y)\)，其中 \(k_m = 2\pi / \lambda_m\)，\(\lambda_m\) 为磁场周期长度。现有一沿 \(z\) 轴以速度 \(v = \beta c\) 运动的惯性系，求在该惯性系中观察到的电磁场。证明当 \(\beta \simeq 1\) 时该电磁场类似于一列频率为 \(\gamma \cdot \beta c k_m\) 的圆偏振电磁波。

在该系中： $$ E_{x}' = \gamma (E_{x} - \beta c B_{y}) = - \gamma \beta c B_{0} \sin k_{m} z $$

\[ E_{y}' = \gamma (E_{y} + \beta c B_{x}) = \gamma \beta c B_{0} \cos k_{m} z \]

\[ E_{z}' = E_{z} = 0 \]

\[ B_{x}' = \gamma (B_{x} + \beta c E_{z}) = \gamma B_{x} = \gamma B_{0} \cos k_{m} z \]

\[ B_{y}' = \gamma (B_{y} - \beta c E_{x}) = \gamma B_{y} = \gamma B_{0} \sin k_{m} z \]

\[ B'_{z} = B_{z} = 0 \]

其中 \(z = \gamma (z' + \beta c t')\) ，得到该波波矢沿 \(- z\) 方向，且圆频率为 \(\gamma \beta c k_{m}\)。

所以\(\beta \approx 1\)时该系中 $$ \vec{B} \approx \frac{\hat{k}}{c} \times \vec{E} $$ 正好为左旋光的电磁场表达式。

6.16 有一无限长均匀带电直线，在其静止参考系中线电荷密度为 \(\lambda\)。该线电荷以速度 \(v = \beta c\) 沿自身长度匀速移动。在与直线相距为 \(d\) 的地方有一以同样速度平行于直线运动的点电荷 \(q\)。分别用下列两种方法求出作用在电荷上的力：

(a) 在带电线静止系中确定力，然后用四维力变换公式；

带电线静止系中： $$ \vec{f} (r) = \frac{\lambda q}{2 \pi \epsilon_{0} d} \hat{r} $$ 使用力的逆变换： $$ \begin{aligned} &\begin{pmatrix} \gamma_{v} f_{x} \\ \gamma_{v} f_{y} \\ \gamma_{v} f_{z} \\ \gamma_{v} \frac{\vec{f} \cdot \vec{v}}{c} \end{pmatrix} = \begin{pmatrix} \gamma & 0 & 0 & \gamma \beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \gamma \beta & 0 & 0 & \gamma \end{pmatrix} \begin{pmatrix} \gamma_{v'} f_{x}' \\ \gamma_{v'} f_{y}' \\ \gamma_{v'} f_{z}' \\ \gamma_{v'} \frac{\vec{f'} \cdot \vec{v'}}{c} \end{pmatrix} \end{aligned} $$

在带电线静止系中，4-力矢量为\(\begin{pmatrix} \frac{\lambda q}{2 \pi \epsilon_{0} d} \hat{r}, 0 \end{pmatrix}\)

所以\(\gamma_{v} \vec{f} = \frac{\lambda q}{2 \pi \epsilon_{0} d} \hat{r}\)

\(\mathbf{F} = \frac{\lambda q}{2 \pi \epsilon_{0} d \gamma} \hat{r}\)

(b) 直接计算线电荷和线电流作用在运动电荷上的电磁力。

在线电荷系中，流矢量为\(\begin{pmatrix} c \lambda , 0 \end{pmatrix}\)，所以地面系中，流矢量变为 $$ \begin{aligned} &\begin{pmatrix} c \lambda_{s} \\ I_{s} \end{pmatrix} = \begin{pmatrix} \gamma & \gamma \beta \\ \gamma \beta & \gamma \end{pmatrix} \begin{pmatrix} c \lambda \\ 0 \end{pmatrix} &= \begin{pmatrix} \gamma c \lambda \\ \gamma \lambda \beta c\end{pmatrix} \end{aligned} $$ 所以离导线 \(d\) 处的电磁场为 \(\vec{E} = \frac{\gamma \lambda}{2 \pi \epsilon_{0} d} \hat{r}\) \(\vec{B} = \frac{\gamma \beta c \lambda}{2 \pi \mu_{0} d} \hat{j} \times \hat{r}\)

此时电荷受力为： $$ \mathbf{F} = \frac{\gamma \lambda q}{2 \pi \epsilon_{0} d} \hat{r} + q \vec{v} \times \frac{\gamma \beta c \lambda}{2 \pi \mu_{0} d} (\hat{j} \times \hat{r}) = \frac{\gamma \lambda q}{2 \pi \epsilon_{0} d} (1 - \frac{v^{2}}{c^{2}}) \hat{r} = \frac{\lambda q}{2 \pi \epsilon_{0} d \gamma} \hat{r} $$

1. 试证, 电磁场的一般变换关系为,

\[ \vec{E}' = \frac{\vec{E} \cdot \vec{v}}{v^2} \vec{v} + \frac{\vec{E} - \frac{\vec{E} \cdot \vec{v}}{v^2} \vec{v} + \vec{v} \times \vec{B}}{\sqrt{1 - \frac{v^2}{c^2}}} \]

\[ \vec{B}' = \frac{\vec{B} \cdot \vec{v}}{v^2} \vec{v} + \frac{\vec{B} - \frac{\vec{B} \cdot \vec{v}}{v^2} \vec{v} - \frac{\vec{v}}{c^2} \times \vec{E}}{\sqrt{1 - \frac{v^2}{c^2}}} \]

由于电磁场张量在变换下满足二阶张量的变换关系，所以对于 \(x\) 方向的Boost： $$ \begin{aligned} F' = \begin{pmatrix} \gamma & 0 & 0 & i\beta\gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -i\beta\gamma & 0 & 0 & \gamma \end{pmatrix} \begin{pmatrix} 0 & B_3 & -B_2 & -iE_1/c \\ -B_3 & 0 & B_1 & -iE_2/c \\ B_2 & -B_1 & 0 & -iE_3/c \\ iE_1/c & iE_2/c & iE_3/c & 0 \end{pmatrix} \begin{pmatrix} \gamma & 0 & 0 & - i\beta\gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ i\beta\gamma & 0 & 0 & \gamma \end{pmatrix} \\ = \begin{pmatrix} 0 & \gamma(B_3 - \tfrac{\beta E_2}{c}) & \gamma(B_2 + \tfrac{\beta E_3}{c}) & -i \frac{E_1}{c} \\ -\gamma(B_3 - \tfrac{\beta E_2}{c}) & 0 & B_1 & -i\gamma(\tfrac{E_2}{c} - \beta B_3) \\ -\gamma(B_2 + \tfrac{\beta E_3}{c}) & -B_1 & 0 & -i\gamma(\tfrac{E_3}{c} + \beta B_2) \\ i \frac{E_1}{c} & i\gamma(\tfrac{E_2}{c} - \beta B_3) & i\gamma(\tfrac{E_3}{c} + \beta B_2) & 0 \end{pmatrix} \end{aligned} $$ 得到对于 Boost 而言 \(E'_{\parallel} = E_{\parallel}\) \(E'_{\perp} = \gamma (E_{\perp} + \vec{v} \times \vec{B})\) \(B'_{\parallel} = B_{\parallel}\) \(B'_{\perp} = \gamma (B_{\perp} - \vec{v} \times \vec{E})\)

所以 $$ \vec{E'} = \vec{E} \cdot \frac{\vec{v}}{v} + \gamma (\vec{E} - \vec{E} \cdot \frac{\vec{v}}{v} + \vec{v} \times \vec{B}) $$

\[ \vec{B'} = \vec{B} \cdot \frac{\vec{v}}{v} + \gamma (\vec{B} - \vec{B} \cdot \frac{\vec{v}}{v} - \frac{\vec{v}}{c^{2}} \times \vec{E}) \]

1. 试证, 匀速运动的点电荷\(q\)产生的电磁场为,

\[ \vec{E} = \frac{q \vec{R}}{4 \pi \epsilon_0 S^3} (1 - \frac{v^2}{c^2}) \]

\[ \vec{B} = \frac{\vec{v}}{c^2} \times \vec{E} \]

其中:

\[ \vec{R} = \vec{r} - \vec{v} t \]

\[ S^2 = (\vec{R} \cdot \frac{\vec{v}}{v})^2 + (1 - \frac{v^2}{c^2}) (\vec{R} - \vec{R} \cdot \frac{\vec{v}}{v^2} \vec{v}) \cdot (\vec{R} - \vec{R} \cdot \frac{\vec{v}}{v^2} \vec{v}) = \frac{(\vec{R} \cdot \vec{v})^2}{c^2} + (1 - \frac{v^2}{c^2}) R^2 \]

证明：在电荷参考系 \(s'\) 中，电磁场为： $$ \vec{E} = \frac{q}{4 \pi \epsilon_{0} r'^{2}} \hat{r'} \qquad \vec{B} = \vec{0} $$ 利用电磁场变换： $$ \vec{E} = \vec{E'} \cdot \frac{\vec{v}}{v} \hat{v} + \gamma (\vec{E'} - \vec{E'} \cdot \frac{\vec{v}}{v} \hat{v}) $$

\[ \vec{B} = \gamma \frac{\vec{v}}{c^{2}} \times \vec{E'} \]

再考虑 \(\vec{E'}\) ，在 \(t\) 时刻，对于场点 \((x,y,z)^{\top}\)， \(S\) 系中相对距离为 $\vec{R} = \vec{r} - \vec{v} t $

变换至 \(S'\)系中，相对距离为(假设电荷运动方向为 \(x\) 方向) \(\vec{r'} = \begin{pmatrix} \gamma (x - vt), y,z \end{pmatrix}^{\top} = \gamma (\vec{R} \cdot \hat{v}) \hat{v} + \vec{R} - (\vec{R} \cdot \hat{v}) \hat{v}\)

所以 \(r'^{2} = \gamma^{2} S^{2}\) $$ \vec{E'} = \frac{q}{4 \pi \epsilon_{0} r'^{3}} \vec{r'} = \frac{q}{4 \pi \epsilon_{0} \gamma^{3} S^{3}} (\gamma (\vec{R} \cdot \hat{v}) \hat{v} + \vec{R} - (\vec{R} \cdot \hat{v}) \hat{v}) $$ 换至 \(S\) 系中 $$ \begin{aligned} \vec{E} &= \vec{E'} \cdot \frac{\vec{v}}{v} \hat{v} + \gamma (\vec{E'} - \vec{E'} \cdot \frac{\vec{v}}{v} \hat{v} ) = \frac{q}{4 \pi \epsilon_{0} \gamma^{3} S^{3}} (\gamma (\vec{R} \cdot \hat{v}) \hat{v} + \gamma (\gamma (\vec{R} \cdot \hat{v}) \hat{v} + \vec{R} - (\vec{R} \cdot \hat{v}) \hat{v} - \gamma (\vec{R} \cdot \hat{v}) \hat{v}) \\ &= \frac{q}{4 \pi \epsilon_{0} \gamma^{3} S^{3}} \gamma \vec{R} = \frac{q \vec{R}}{4 \pi \epsilon_0 S^3} (1 - \frac{v^2}{c^2}) \end{aligned} $$

\[ \vec{B} = \gamma \frac{\vec{v}}{c^{2}} \times \vec{E'} = \gamma \frac{\vec{v}}{c^{2}} \times \frac{q}{4 \pi \epsilon_{0} \gamma^{3} S^{3}} (\gamma (\vec{R} \cdot \hat{v}) \hat{v} + \vec{R} - (\vec{R} \cdot \hat{v}) \hat{v}) = \frac{\vec{v}}{c^{2}} \times \frac{q}{4 \pi \epsilon_{0} \gamma^{3} S^{3}} \gamma \vec{R} = \frac{\vec{v}}{c^2} \times \vec{E} \]