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 $\phi ,A;E,B$。

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：

$${\varphi }^{\prime }=\frac{1}{4\pi {ϵ}_0}\frac{\overrightarrow{r^{\prime }}\cdot \overrightarrow{p_0}}{r^{\prime 3}}$$$${\overrightarrow{A}}^{\prime }=0$$

换至地面系中：

$$\left(\begin{array}{c}{A}_x\\ A_y\\ A_z\\ \frac{\varphi }{c}\end{array}\right)=\left(\begin{array}{cccc}\gamma & 0& 0& \gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ \gamma \beta & 0& 0& \gamma \end{array}\right)\left(\begin{array}{c}{A}_x^{\prime }\\ A_y^{\prime }\\ A_z^{\prime }\\ \frac{{\varphi }^{\prime }}{c}\end{array}\right)=\left(\begin{array}{cccc}\gamma & 0& 0& \gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ \gamma \beta & 0& 0& \gamma \end{array}\right)\left(\begin{array}{c}0\\ 0\\ 0\\ \frac{1}{4\pi {ϵ}_{0}c}\frac{\overrightarrow{r^{\prime }}\cdot \overrightarrow{p_0}}{r^{\prime 3}}\end{array}\right)=\left(\begin{array}{c}\frac{\gamma \beta }{4\pi {ϵ}_{0}c}\frac{\overrightarrow{r^{\prime }}\cdot \overrightarrow{p_0}}{r^{\prime 3}}\\ 0\\ 0\\ \frac{\gamma }{4\pi {ϵ}_{0}c}\frac{\overrightarrow{r^{\prime }}\cdot \overrightarrow{p_0}}{r^{\prime 3}}\end{array}\right)$$

所以$\varphi =\frac{\gamma }{4\pi {ϵ}_0}\frac{\overrightarrow{r^{\prime }}\cdot \overrightarrow{p_0}}{r^{\prime 3}}$， $\overrightarrow{A}={\left(\begin{array}{c}\frac{\gamma \beta }{4\pi {ϵ}_{0}c}\frac{\overrightarrow{r^{\prime }}\cdot \overrightarrow{p_0}}{r^{\prime 3}},0,0\end{array}\right)}^{\mathrm{\top }}$

其中$\overrightarrow{r^{\prime }}={\left(\begin{array}{c}\gamma (x-\beta ct),y,z\end{array}\right)}^{\mathrm{\top }}$

在电偶极子系中，电偶极子产生的电场为：

$$\overrightarrow{E}=\frac{1}{4\pi {ϵ}_0}\frac{3(\overrightarrow{p_0}\cdot \overrightarrow{r^{\prime }})\overrightarrow{r^{\prime }}-r^{\prime 2}\overrightarrow{p_0}}{r^{\prime 5}}$$

换至地面系中，得到：

$$E_x=E_x^{\prime }=\frac{1}{4\pi {ϵ}_0}\frac{3(\overrightarrow{p_0}\cdot \overrightarrow{r^{\prime }})x^{\prime }-r^{\prime 2}(\overrightarrow{p_0}\cdot \widehat{x^{\prime }})}{r^{\prime 5}}$$$$E_y=\gamma (E_y^{\prime }+\beta c{B}_z^{\prime })=\gamma E_y^{\prime }=\frac{\gamma }{4\pi {ϵ}_0}\frac{3(\overrightarrow{p_0}\cdot \overrightarrow{r^{\prime }})y^{\prime }-r^{\prime 2}(\overrightarrow{p_0}\cdot \widehat{y^{\prime }})}{r^{\prime 5}}$$$$E_z=\gamma (E_z^{\prime }-\beta c{B}_y^{\prime })=\gamma E_z^{\prime }=\frac{\gamma }{4\pi {ϵ}_0}\frac{3(\overrightarrow{p_0}\cdot \overrightarrow{r^{\prime }})z^{\prime }-r^{\prime 2}(\overrightarrow{p_0}\cdot \widehat{z^{\prime }})}{r^{\prime 5}}$$$$B_x=B_x^{\prime }=0$$$$B_y=\gamma (B_{y^{\prime }}-\beta \frac{E_z^{\prime }}{c})=-\frac{\gamma }{\beta c}{E}_z^{\prime }=-\frac{\gamma }{4\pi {ϵ}_0\beta c}\frac{3(\overrightarrow{p_0}\cdot \overrightarrow{r^{\prime }})z^{\prime }-r^{\prime 2}(\overrightarrow{p_0}\cdot \widehat{z^{\prime }})}{r^{\prime 5}}$$$$B_z=\gamma (B_{z^{\prime }}+\beta \frac{E_y^{\prime }}{c})=\frac{\gamma }{\beta c}{E}_y^{\prime }=\frac{\gamma }{4\pi {ϵ}_0\beta c}\frac{3(\overrightarrow{p_0}\cdot \overrightarrow{r^{\prime }})y^{\prime }-r^{\prime 2}(\overrightarrow{p_0}\cdot \widehat{y^{\prime }})}{r^{\prime 5}}$$

其中$\overrightarrow{r^{\prime }}={\left(\begin{array}{c}\gamma (x-\beta ct),y,z\end{array}\right)}^{\mathrm{\top }}$

6.15 有一沿 $z$ 轴方向螺旋进动的静磁场 $\overrightarrow{B}=B_0(\cos k_{m}z{\hat{e}}_x+\sin k_{m}z{\hat{e}}_y)$，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：

$$E_x^{\prime }=\gamma (E_x-\beta c{B}_y)=-\gamma \beta c{B}_0\sin k_{m}z$$$$E_y^{\prime }=\gamma (E_y+\beta c{B}_x)=\gamma \beta c{B}_0\cos k_{m}z$$$$E_z^{\prime }=E_z=0$$$$B_x^{\prime }=\gamma (B_x+\beta c{E}_z)=\gamma B_x=\gamma B_0\cos k_{m}z$$$$B_y^{\prime }=\gamma (B_y-\beta c{E}_x)=\gamma B_y=\gamma B_0\sin k_{m}z$$$$B_z^{\prime }=B_z=0$$

其中 $z=\gamma (z^{\prime }+\beta c{t}^{\prime })$ ，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

$$\overrightarrow{B}\approx \frac{\hat{k}}{c}\times \overrightarrow{E}$$

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：

$$\overrightarrow{f}(r)=\frac{\lambda q}{2\pi {ϵ}_{0}d}\hat{r}$$

使用力的逆变换：

$$\left(\begin{array}{c}{\gamma }_v{f}_x\\ {\gamma }_v{f}_y\\ {\gamma }_v{f}_z\\ {\gamma }_v\frac{\overrightarrow{f}\cdot \overrightarrow{v}}{c}\end{array}\right)=\left(\begin{array}{cccc}\gamma & 0& 0& \gamma \beta \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ \gamma \beta & 0& 0& \gamma \end{array}\right)\left(\begin{array}{c}{\gamma }_{v^{\prime }}{f}_x^{\prime }\\ {\gamma }_{v^{\prime }}{f}_y^{\prime }\\ {\gamma }_{v^{\prime }}{f}_z^{\prime }\\ {\gamma }_{v^{\prime }}\frac{\overrightarrow{f^{\prime }}\cdot \overrightarrow{v^{\prime }}}{c}\end{array}\right)$$

在带电线静止系中，4-力矢量为$\left(\begin{array}{c}\frac{\lambda q}{2\pi {ϵ}_{0}d}\hat{r},0\end{array}\right)$

所以${\gamma }_v\overrightarrow{f}=\frac{\lambda q}{2\pi {ϵ}_{0}d}\hat{r}$

$\mathbf{F}=\frac{\lambda q}{2\pi {ϵ}_{0}d\gamma }\hat{r}$

(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$\left(\begin{array}{c}c\lambda ,0\end{array}\right)$，所以地面系中，流矢量变为

$$\left(\begin{array}{c}c{\lambda }_s\\ I_s\end{array}\right)=\left(\begin{array}{cc}\gamma & \gamma \beta \\ \gamma \beta & \gamma \end{array}\right)\left(\begin{array}{c}c\lambda \\ 0\end{array}\right)=\left(\begin{array}{c}\gamma c\lambda \\ \gamma \lambda \beta c\end{array}\right)$$

所以离导线 $d$ 处的电磁场为 $\overrightarrow{E}=\frac{\gamma \lambda }{2\pi {ϵ}_{0}d}\hat{r}$ $\overrightarrow{B}=\frac{\gamma \beta c\lambda }{2\pi {\mu }_{0}d}\hat{j}\times \hat{r}$

此时电荷受力为：

$$\mathbf{F}=\frac{\gamma \lambda q}{2\pi {ϵ}_{0}d}\hat{r}+q\overrightarrow{v}\times \frac{\gamma \beta c\lambda }{2\pi {\mu }_{0}d}(\hat{j}\times \hat{r})=\frac{\gamma \lambda q}{2\pi {ϵ}_{0}d}(1-\frac{v^2}{c^2})\hat{r}=\frac{\lambda q}{2\pi {ϵ}_{0}d\gamma }\hat{r}$$

2. 试证, 电磁场的一般变换关系为,

$${\overrightarrow{E}}^{\prime }=\frac{\overrightarrow{E}\cdot \overrightarrow{v}}{v^2}\overrightarrow{v}+\frac{\overrightarrow{E}-\frac{\overrightarrow{E}\cdot \overrightarrow{v}}{v^2}\overrightarrow{v}+\overrightarrow{v}\times \overrightarrow{B}}{\sqrt{1-\frac{v^2}{c^2}}}$$$${\overrightarrow{B}}^{\prime }=\frac{\overrightarrow{B}\cdot \overrightarrow{v}}{v^2}\overrightarrow{v}+\frac{\overrightarrow{B}-\frac{\overrightarrow{B}\cdot \overrightarrow{v}}{v^2}\overrightarrow{v}-\frac{\overrightarrow{v}}{c^2}\times \overrightarrow{E}}{\sqrt{1-\frac{v^2}{c^2}}}$$

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：

$$\begin{gathered} F^{\prime }=\left(\begin{array}{cccc}\gamma & 0& 0& i\beta \gamma \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ -i\beta \gamma & 0& 0& \gamma \end{array}\right)\left(\begin{array}{cccc}0& B_3& -B_2& -i{E}_1/c\\ -B_3& 0& B_1& -i{E}_2/c\\ B_2& -B_1& 0& -i{E}_3/c\\ i{E}_1/c& i{E}_2/c& i{E}_3/c& 0\end{array}\right)\left(\begin{array}{cccc}\gamma & 0& 0& -i\beta \gamma \\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ i\beta \gamma & 0& 0& \gamma \end{array}\right) \\ =\left(\begin{array}{cccc}0& \gamma (B_3-\frac{\beta E_2}{c})& \gamma (B_2+\frac{\beta E_3}{c})& -i\frac{E_1}{c}\\ -\gamma (B_3-\frac{\beta E_2}{c})& 0& B_1& -i\gamma (\frac{E_2}{c}-\beta B_3)\\ -\gamma (B_2+\frac{\beta E_3}{c})& -B_1& 0& -i\gamma (\frac{E_3}{c}+\beta B_2)\\ i\frac{E_1}{c}& i\gamma (\frac{E_2}{c}-\beta B_3)& i\gamma (\frac{E_3}{c}+\beta B_2)& 0\end{array}\right) \end{gathered}$$

得到对于 Boost 而言 $E_{\parallel }^{\prime }=E_{\parallel }$ $E_{\perp }^{\prime }=\gamma (E_{\perp }+\overrightarrow{v}\times \overrightarrow{B})$ $B_{\parallel }^{\prime }=B_{\parallel }$ $B_{\perp }^{\prime }=\gamma (B_{\perp }-\overrightarrow{v}\times \overrightarrow{E})$

所以

$$\overrightarrow{E^{\prime }}=\overrightarrow{E}\cdot \frac{\overrightarrow{v}}{v}+\gamma (\overrightarrow{E}-\overrightarrow{E}\cdot \frac{\overrightarrow{v}}{v}+\overrightarrow{v}\times \overrightarrow{B})$$$$\overrightarrow{B^{\prime }}=\overrightarrow{B}\cdot \frac{\overrightarrow{v}}{v}+\gamma (\overrightarrow{B}-\overrightarrow{B}\cdot \frac{\overrightarrow{v}}{v}-\frac{\overrightarrow{v}}{c^2}\times \overrightarrow{E})$$

3. 试证, 匀速运动的点电荷$q$产生的电磁场为,

$$\overrightarrow{E}=\frac{q\overrightarrow{R}}{4\pi {ϵ}_0{S}^3}(1-\frac{v^2}{c^2})$$$$\overrightarrow{B}=\frac{\overrightarrow{v}}{c^2}\times \overrightarrow{E}$$

其中:

$$\overrightarrow{R}=\overrightarrow{r}-\overrightarrow{v}t$$$$S^2=(\overrightarrow{R}\cdot \frac{\overrightarrow{v}}{v}{)}^2+(1-\frac{v^2}{c^2})(\overrightarrow{R}-\overrightarrow{R}\cdot \frac{\overrightarrow{v}}{v^2}\overrightarrow{v})\cdot (\overrightarrow{R}-\overrightarrow{R}\cdot \frac{\overrightarrow{v}}{v^2}\overrightarrow{v})=\frac{(\overrightarrow{R}\cdot \overrightarrow{v}{)}^2}{c^2}+(1-\frac{v^2}{c^2})R^2$$

证明：在电荷参考系 $s^{\prime }$ 中，电磁场为：

$$\overrightarrow{E}=\frac{q}{4\pi {ϵ}_0{r}^{\prime 2}}\widehat{r^{\prime }}\overrightarrow{B}=\overrightarrow{0}$$

利用电磁场变换：

$$\overrightarrow{E}=\overrightarrow{E^{\prime }}\cdot \frac{\overrightarrow{v}}{v}\hat{v}+\gamma (\overrightarrow{E^{\prime }}-\overrightarrow{E^{\prime }}\cdot \frac{\overrightarrow{v}}{v}\hat{v})$$$$\overrightarrow{B}=\gamma \frac{\overrightarrow{v}}{c^2}\times \overrightarrow{E^{\prime }}$$

再考虑 $\overrightarrow{E^{\prime }}$ ，在 $t$ 时刻，对于场点 $(x,y,z{)}^{\mathrm{\top }}$， $S$ 系中相对距离为 $\overrightarrow{R}=\overrightarrow{r}-\overrightarrow{v}t$

变换至 $S^{\prime }$系中，相对距离为(假设电荷运动方向为 $x$ 方向) $\overrightarrow{r^{\prime }}={\left(\begin{array}{c}\gamma (x-vt),y,z\end{array}\right)}^{\mathrm{\top }}=\gamma (\overrightarrow{R}\cdot \hat{v})\hat{v}+\overrightarrow{R}-(\overrightarrow{R}\cdot \hat{v})\hat{v}$

所以 $r^{\prime 2}={\gamma }^2{S}^2$

$$\overrightarrow{E^{\prime }}=\frac{q}{4\pi {ϵ}_0{r}^{\prime 3}}\overrightarrow{r^{\prime }}=\frac{q}{4\pi {ϵ}_0{\gamma }^3{S}^3}(\gamma (\overrightarrow{R}\cdot \hat{v})\hat{v}+\overrightarrow{R}-(\overrightarrow{R}\cdot \hat{v})\hat{v})$$

换至 $S$ 系中

$$\begin{array}{c}\overrightarrow{E}=\overrightarrow{E^{\prime }}\cdot \frac{\overrightarrow{v}}{v}\hat{v}+\gamma (\overrightarrow{E^{\prime }}-\overrightarrow{E^{\prime }}\cdot \frac{\overrightarrow{v}}{v}\hat{v})=\frac{q}{4\pi {ϵ}_0{\gamma }^3{S}^3}(\gamma (\overrightarrow{R}\cdot \hat{v})\hat{v}+\gamma (\gamma (\overrightarrow{R}\cdot \hat{v})\hat{v}+\overrightarrow{R}-(\overrightarrow{R}\cdot \hat{v})\hat{v}-\gamma (\overrightarrow{R}\cdot \hat{v})\hat{v})\\ =\frac{q}{4\pi {ϵ}_0{\gamma }^3{S}^3}\gamma \overrightarrow{R}=\frac{q\overrightarrow{R}}{4\pi {ϵ}_0{S}^3}(1-\frac{v^2}{c^2})\end{array}$$$$\overrightarrow{B}=\gamma \frac{\overrightarrow{v}}{c^2}\times \overrightarrow{E^{\prime }}=\gamma \frac{\overrightarrow{v}}{c^2}\times \frac{q}{4\pi {ϵ}_0{\gamma }^3{S}^3}(\gamma (\overrightarrow{R}\cdot \hat{v})\hat{v}+\overrightarrow{R}-(\overrightarrow{R}\cdot \hat{v})\hat{v})=\frac{\overrightarrow{v}}{c^2}\times \frac{q}{4\pi {ϵ}_0{\gamma }^3{S}^3}\gamma \overrightarrow{R}=\frac{\overrightarrow{v}}{c^2}\times \overrightarrow{E}$$