费曼物理学（2） 第6次作业

Chasse_neige

1 运动电荷的场 (25-5)

直接计算匀速运动电荷产生的电磁场，并将其与静止电荷电磁场洛伦兹变换后的结果做比较。

假设电荷以速度$v$ 朝$x$ 轴正方向运动，假设 $t=0$ 时电荷正好在原点，计算 $t=0$ 时刻空间的电场分布

考虑李纳-维谢尔势（用${\overrightarrow{r}}^{\prime }=\overrightarrow{r}+\overrightarrow{v}t$ 表示推迟后的相对位矢）

$$|\overrightarrow{r}+\overrightarrow{v}t|=ct$$$$t=\frac{|\overrightarrow{r^{\prime }}-\overrightarrow{r}|}{v}=\frac{r^{\prime }}{c}$$$$r^{\prime 2}+r^2-2\overrightarrow{r^{\prime }}\cdot \overrightarrow{r}={\beta }^2{r}^{\prime 2}$$$$\frac{\overrightarrow{r^{\prime }}-\overrightarrow{r}}{|\overrightarrow{r^{\prime }}-\overrightarrow{r}|}\cdot \overrightarrow{r}=\overrightarrow{r}\cdot \hat{x}$$$$\overrightarrow{r^{\prime }}\cdot \overrightarrow{r}-r^2=\beta r^{\prime }\overrightarrow{r}\cdot \hat{x}$$$$r^2+(1-{\beta }^2)r^{\prime 2}-2\beta r^{\prime }\overrightarrow{r}\cdot \hat{x}-2r^2=0$$$$(1-{\beta }^2)r^{\prime 2}-2\beta (\overrightarrow{r}\cdot \hat{x})r^{\prime }-r^2=0$$$$r^{\prime }=\frac{\beta (\overrightarrow{r}\cdot \hat{x})+\sqrt{{\beta }^2(\overrightarrow{r}\cdot \hat{x}{)}^2+(1-{\beta }^2)r^2}}{1-{\beta }^2}$$$$r^{\prime }-\frac{\overrightarrow{v}\cdot \overrightarrow{r^{\prime }}}{c}=ct(1-{\beta }^2)-\frac{\overrightarrow{r}\cdot \overrightarrow{v}}{c}=r^{\prime }(1-{\beta }^2)-\frac{\overrightarrow{r}\cdot \overrightarrow{v}}{c}=\sqrt{{\beta }^2(\overrightarrow{r}\cdot \hat{x}{)}^2+(1-{\beta }^2)r^2}$$$$\begin{array}{c}\varphi =\frac{q}{4\pi {ϵ}_0(r^{\prime }-\frac{\overrightarrow{v}\cdot \overrightarrow{r^{\prime }}}{c})}\\ \overrightarrow{A}=\varphi \overrightarrow{v}=\frac{q\overrightarrow{v}}{4\pi {ϵ}_0{c}^2(r^{\prime }-\frac{\overrightarrow{v}\cdot \overrightarrow{r^{\prime }}}{c})}\end{array}$$

所以

$$\begin{array}{c}\overrightarrow{E}=-\mathrm{\nabla }\varphi -\frac{\partial }{\partial t}\overrightarrow{A}=-\mathrm{\nabla }\frac{q}{4\pi {ϵ}_0(r^{\prime }-\frac{\overrightarrow{v}\cdot \overrightarrow{r^{\prime }}}{c})}-\frac{\partial }{\partial t}\frac{q\overrightarrow{v}}{4\pi {ϵ}_0{c}^2(r^{\prime }-\frac{\overrightarrow{v}\cdot \overrightarrow{r^{\prime }}}{c})}\\ =-\frac{q}{4\pi {ϵ}_0}(-\frac{2{\beta }^2(\overrightarrow{r}\cdot \hat{x})\hat{x}+2(1-{\beta }^2)\overrightarrow{r}}{2{\sqrt{{\beta }^2(\overrightarrow{r}\cdot \hat{x}{)}^2+(1-{\beta }^2)r^2}}^3}+\frac{2{\beta }^2(\overrightarrow{r}\cdot \hat{x})\hat{x}+2(1-{\beta }^2)\overrightarrow{r}}{2{\sqrt{{\beta }^2(\overrightarrow{r}\cdot \hat{x}{)}^2+(1-{\beta }^2)r^2}}^3}\cdot \frac{\overrightarrow{v}}{c^2}\overrightarrow{v})\\ =\frac{q}{4\pi {ϵ}_0}\frac{(1-{\beta }^2)\overrightarrow{r}}{({\beta }^2(\overrightarrow{r}\cdot \hat{v}{)}^2+(1-{\beta }^2)r^2{)}^{\frac{3}{2}}}\end{array}$$

对比洛伦兹变换结果

在粒子随动系中

$$\overrightarrow{E^{\prime }}=\frac{q}{4\pi {ϵ}_0{r}^{\prime 3}}\overrightarrow{r^{\prime }}$$

换系后

$$\overrightarrow{E}=\frac{q}{4\pi {ϵ}_0}(-(\gamma -1)\frac{\overrightarrow{r^{\prime }}\cdot \hat{v}}{r^{\prime 3}}\hat{v}+\gamma \frac{\overrightarrow{r^{\prime }}}{r^{\prime 3}})$$

带入 $t=0$ 时

$$\overrightarrow{r^{\prime }}=(\gamma -1)(\overrightarrow{r}\cdot \hat{v})\hat{v}+\overrightarrow{r}$$

所以

$$\begin{array}{c}\overrightarrow{E}=\frac{q}{4\pi {ϵ}_0}\frac{-(\gamma -1)\gamma (\overrightarrow{r}\cdot \hat{v})\hat{v}+\gamma (\gamma -1)(\overrightarrow{r}\cdot \hat{v})\hat{v}+\gamma \overrightarrow{r}}{{\left(\frac{{\beta }^2(\overrightarrow{r}\cdot \hat{v}{)}^2+(1-{\beta }^2)r^2}{1-{\beta }^2}\right)}^{\frac{3}{2}}}\\ =\frac{q}{4\pi {ϵ}_0}\frac{\gamma \overrightarrow{r}}{{\left(\frac{{\beta }^2(\overrightarrow{r}\cdot \hat{v}{)}^2+(1-{\beta }^2)r^2}{1-{\beta }^2}\right)}^{\frac{3}{2}}}\\ =\frac{q}{4\pi {ϵ}_0}\frac{(1-{\beta }^2)\overrightarrow{r}}{({\beta }^2(\overrightarrow{r}\cdot \hat{v}{)}^2+(1-{\beta }^2)r^2{)}^{\frac{3}{2}}}\end{array}$$

可以看出，两种计算方式的结果是相同的

2 麦克斯韦方程组的洛伦兹协变性 (26-3)

请证明麦克斯韦方程组在洛伦兹变换下保持协变，即麦克斯韦方程组形式在洛伦兹变换下保持不变。

麦克斯韦方程组在张量形式下可以表示为：

$$\begin{array}{rl}{\partial }_{\mu }{F}^{\mu \nu }& =\frac{4\pi }{c}{j}^{\nu }\\ {\partial }_{\mu }{}^{\star }{F}^{\mu \nu }& =0\end{array}$$

电磁场张量 $F^{\mu \nu }$ 定义为：

$$F^{\mu \nu }=\left(\begin{array}{cccc}0& -E_x& -E_y& -E_z\\ E_x& 0& -B_z& B_y\\ E_y& B_z& 0& -B_x\\ E_z& -B_y& B_x& 0\end{array}\right).$$

对偶电磁场张量 ${}^{\star }{F}^{\mu \nu }$ 定义为：

$${}^{\star }{F}^{\mu \nu }=\left(\begin{array}{cccc}0& B_x& B_y& B_z\\ -B_x& 0& E_z& -E_y\\ -B_y& -E_z& 0& E_x\\ -B_z& E_y& -E_x& 0\end{array}\right).$$

四维流密度 $j^{\nu }$ 定义为：

$$j^{\nu }=(c\rho ,\overrightarrow{J})$$

洛伦兹变换可以表示为：

$$x^{\prime \mu }={\mathrm{\Lambda }}^{\mu }{}_{\nu }{x}^{\nu }$$

其中 ${\mathrm{\Lambda }}^{\mu }{}_{\nu }$ 是洛伦兹变换矩阵，满足：

$${\mathrm{\Lambda }}^{\mu }{}_{\alpha }{\mathrm{\Lambda }}^{\nu }{}_{\beta }{\eta }_{\mu \nu }={\eta }_{\alpha \beta },$$

其中 ${\eta }_{\mu \nu }=\text{diag}(-1,1,1,1)$

$$F^{\prime \mu \nu }={\mathrm{\Lambda }}^{\mu }{}_{\alpha }{\mathrm{\Lambda }}^{\nu }{}_{\beta }{F}^{\alpha \beta },$$$$j^{\prime \nu }={\mathrm{\Lambda }}^{\nu }{}_{\beta }{j}^{\beta }$$$${\partial }_{\mu }^{\prime }=\frac{\partial }{\partial x^{\prime \mu }}=\frac{\partial x^{\alpha }}{\partial x^{\prime \mu }}\frac{\partial }{\partial x^{\alpha }}={\mathrm{\Lambda }}^{\alpha }{}_{\mu }{\partial }_{\alpha }$$$$F^{\prime \mu \nu }={\mathrm{\Lambda }}^{\mu }{}_{\alpha }{\mathrm{\Lambda }}^{\nu }{}_{\beta }{F}^{\alpha \beta }$$$$j^{\prime \nu }={\mathrm{\Lambda }}^{\nu }{}_{\beta }{j}^{\beta }$$$${\partial }_{\mu }^{\prime }{F}^{\prime \mu \nu }={\mathrm{\Lambda }}^{\delta }{}_{\mu }{\partial }_{\delta }\left({\mathrm{\Lambda }}^{\mu }{}_{\alpha }{\mathrm{\Lambda }}^{\nu }{}_{\beta }{F}^{\alpha \beta }\right)$$$$\begin{array}{c}{\partial }_{\mu }^{\prime }{F}^{\prime \mu \nu }={\delta }^{\delta }{}_{\alpha }{\mathrm{\Lambda }}^{\nu }{}_{\beta }{\partial }_{\delta }{F}^{\alpha \beta }\\ ={\mathrm{\Lambda }}^{\nu }{}_{\beta }{\partial }_{\alpha }{F}^{\alpha \beta }=\frac{4\pi }{c}{\mathrm{\Lambda }}^{\nu }{}_{\beta }{j}^{\beta }=\frac{4\pi }{c}{j}^{\prime \nu }\end{array}$$

同理，对于对偶电磁场张量：

$${\partial }_{\mu }^{\prime }{}^{\star }{F}^{\prime \mu \nu }=0$$

所以麦克斯韦方程组是洛伦兹协变的。

3 对称性与电磁动量 (27-6)

在课堂上，大家表示电磁场能量-动量张量的推导比较繁琐，并且看不出太明显的物理意义。在本题当中，我们尝试从时空平移不变性出发，利用诺特定理推导能量-动量张量的表达式。从而将能量-动量张量与对称性联系起来。

(1). 考虑一个无穷小的变换: $x^{\mu }\to x^{\mu }+{ϵ}^{\mu }(x)$。在此无穷小变换下，电磁势按照如下变换: $A^{\mu }\to A^{\mu }+{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }$。对于一般的理论而言，其作用量只明显包含电磁势 $A^{\mu }$ 及其导数 ${\partial }_{\nu }{A}^{\mu }$。请证明在上述变换下，作用量的变化满足:

$$\delta S=\int d^{4}x(\frac{\partial \mathcal{L}}{\partial A^{\mu }}{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }+\frac{\partial \mathcal{L}}{\partial ({\partial }_{\sigma }{A}^{\mu })}{\partial }_{\sigma }[{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }])$$

证明

$$\begin{array}{c}{A}^{\mu }\to A^{\mu }+{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }\\ {\partial }_{\sigma }{A}^{\mu }\to {\partial }_{\sigma }{A}^{\mu }+{\partial }_{\sigma }{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }\end{array}$$

所以

$$\begin{array}{c}\delta S=\frac{\partial S}{\partial A^{\mu }}{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }+\frac{\partial S}{\partial ({\partial }_{\sigma }{A}^{\mu })}{\partial }_{\sigma }{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }\\ =\int d^{4}x(\frac{\partial \mathcal{L}}{\partial A^{\mu }}{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }+\frac{\partial \mathcal{L}}{\partial ({\partial }_{\sigma }{A}^{\mu })}{\partial }_{\sigma }[{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }])\end{array}$$

(2). 在上述的基础上，请证明作用量的变化满足:

$$\delta S=\int d^{4}x{T}^{\mu \nu }{\partial }_{\mu }{ϵ}_{\nu }(x),T^{\mu \nu }=\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}_{\sigma })}{\partial }^{\nu }{A}_{\sigma }-{\eta }^{\mu \nu }\mathcal{L}$$

证明

$$\begin{array}{c}\delta S=\int d^{4}x(\frac{\partial \mathcal{L}}{\partial A^{\mu }}{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }+\frac{\partial \mathcal{L}}{\partial ({\partial }_{\sigma }{A}^{\mu })}{\partial }_{\sigma }[{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\mu }])\\ =\int d^{4}x(\frac{\partial \mathcal{L}}{\partial A^{\mu }}{ϵ}^{\nu }(x)+\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}^{\sigma })}{\partial }_{\mu }[{ϵ}^{\nu }(x){\partial }_{\nu }{A}^{\sigma }])\\ =\int d^{4}x(\frac{\partial \mathcal{L}}{\partial A^{\mu }}{ϵ}^{\nu }(x)+\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}^{\sigma })}{\partial }_{\mu }({\partial }_{\nu }{A}^{\sigma }){ϵ}^{\nu }(x)+\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}^{\sigma })}{\partial }_{\nu }{A}^{\sigma }{\partial }_{\mu }{ϵ}^{\nu }(x))\\ =\int d^{4}x({\partial }_{\nu }\mathcal{L}{ϵ}^{\nu }(x)+\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}^{\sigma })}{\partial }_{\nu }{A}^{\sigma }{\partial }_{\mu }{ϵ}^{\nu }(x))\\ =\int d^{4}x(\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}^{\sigma })}{\partial }_{\nu }{A}^{\sigma }{\partial }_{\mu }{ϵ}^{\nu }(x)-\mathcal{L}{\partial }_{\nu }{ϵ}^{\nu }(x))\\ =\int d^{4}x(\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}_{\sigma })}{\partial }^{\nu }{A}_{\sigma }{\partial }_{\mu }{ϵ}_{\nu }(x)-\mathcal{L}{\partial }_{\mu }{\eta }^{\mu \nu }{ϵ}_{\nu }(x))\\ =\int d^{4}x{T}^{\mu \nu }{\partial }_{\mu }{ϵ}_{\nu }(x),T^{\mu \nu }=\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}_{\sigma })}{\partial }^{\nu }{A}_{\sigma }-{\eta }^{\mu \nu }\mathcal{L}\end{array}$$

(3). 对于一般的理论而言，如果其满足时空平移不变性，即 ${ϵ}^{\nu }(x)={ϵ}^{\nu }$ 为一个不依赖于时空坐标的函数时，体系的作用量应该在该时空平移下保持不变: $\delta S=0$。在分部积分的意义下，这意味着我们有 ${\partial }_{\nu }{T}^{\mu \nu }=0$。对于电磁理论来说，其作用量可以写为:

$$S=-\int d^{4}x\left(\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }\right)$$

请利用第二问的结论证明电磁理论的能量-动量张量满足:

$$T^{\mu \nu }=-F^{\mu \sigma }{\partial }_{\nu }{A}^{\sigma }-{\eta }^{\mu \nu }\mathcal{L}$$$$\mathcal{L}=-\frac{1}{4}({\partial }_{\mu }{A}_{\sigma }-{\partial }_{\sigma }{A}_{\mu })({\partial }^{\mu }{A}^{\sigma }-{\partial }^{\sigma }{A}^{\mu })=-\frac{1}{2}{\partial }_{\mu }{A}_{\sigma }{\partial }^{\mu }{A}^{\sigma }+\frac{1}{2}{\partial }_{\mu }{A}_{\sigma }{\partial }^{\sigma }{A}^{\mu }$$$$\begin{array}{c}{T}^{\mu \nu }=\frac{\partial \mathcal{L}}{\partial ({\partial }_{\mu }{A}_{\sigma })}{\partial }^{\nu }{A}_{\sigma }-{\eta }^{\mu \nu }\mathcal{L}\\ =(-{\partial }^{\mu }{A}^{\sigma }+{\partial }^{\sigma }{A}^{\mu }){\partial }^{\nu }{A}_{\sigma }-{\eta }^{\mu \nu }\mathcal{L}\\ =-F^{\mu \sigma }{\partial }_{\nu }{A}^{\sigma }-{\eta }^{\mu \nu }\mathcal{L}\end{array}$$

(4) 对于一般的电磁理论而言，我们可以采用希尔伯特的定义方式，让其能量-动量张量是对称张量。具体而言，对于电磁理论而言，其能量-动量张量对称化之后可以写为:

$$T^{\mu \nu }={\eta }^{\sigma \lambda }{F}^{\mu \sigma }{F}^{\lambda \nu }-{\eta }^{\mu \nu }\mathcal{L}$$

请证明能量-动量张量零分量的守恒方程即为电磁场的守恒方程:

$${\partial }_{\mu }{T}^{\mu 0}=0,\Rightarrow \frac{1}{2}\frac{\partial }{\partial t}(\overrightarrow{E}\cdot \overrightarrow{E}+\overrightarrow{B}\cdot \overrightarrow{B})+\mathrm{\nabla }\cdot (\overrightarrow{E}\times \overrightarrow{B})=0$$

证明

$$\begin{array}{c}{\partial }_{\mu }{T}^{\mu 0}={\partial }_{\mu }({\eta }^{\sigma \lambda }{F}^{\mu \sigma }{F}^{\lambda 0}-{\eta }^{\mu 0}\mathcal{L})\\ ={\partial }_{\mu }(-(F^{\mu 1}{F}^{10}+F^{\mu 2}{F}^{20}+F^{\mu 3}{F}^{30})+F^{\mu 0}{F}^{00})+{\partial }_0\mathcal{L}\\ ={\partial }_0(E^2)+{\partial }_i(F^{ij}{F}^{j0})-\frac{\partial }{\partial t}\mathcal{L}\\ =\frac{\partial }{\partial t}{E}^2+{\partial }_i({ϵ}^{ijk}{E}_j{B}_k)-\frac{\partial }{\partial t}(\frac{1}{2}(E^2-B^2))\\ =\frac{1}{2}\frac{\partial }{\partial t}(\overrightarrow{E}\cdot \overrightarrow{E}+\overrightarrow{B}\cdot \overrightarrow{B})+\mathrm{\nabla }\cdot (\overrightarrow{E}\times \overrightarrow{B})=0\end{array}$$

4 菲涅尔定律及其应用 (33-1)

菲涅尔棱镜是一种将线性偏振光转换为圆偏振光的光学器件。如图所示，光线垂直线入射到棱镜的表面，然后在棱镜内部经历两次全内反射，最后再垂直射出棱镜。光线在棱镜内每次反射都会使平行于入射平面的光波分量与垂直于入射平面的光波分量之间产生 45° 的相位差。因此，经过两次内反射后，原本与入射平面成 45° 线偏振的光波会变成圆偏振光。

(1). 对于偏振方向平行入射平面的电磁波而言，请计算单次内部反射后，反射波与入射波的相位差。

$${\delta }_p=2\arctan \frac{n}{\cos {\theta }_i}\sqrt{(n\sin {\theta }_i{)}^2-1}$$

(2). 当电磁波的偏振方向垂直于入射平面时，请计算反射波相对于入射波的相位差。

$${\delta }_s=2\arctan \frac{1}{n\cos {\theta }_i}\sqrt{(n\sin {\theta }_i{)}^2-1}$$

(3). 如果菲涅尔棱镜中的两次全内反射的入射角均为 ${\theta }_i={53.3}^{\circ }$，请计算菲涅尔棱镜相对于周围环境的折射率。

$$2(\arctan \frac{n}{\cos {\theta }_i}\sqrt{(n\sin {\theta }_i{)}^2-1}-\arctan \frac{1}{n\cos {\theta }_i}\sqrt{(n\sin {\theta }_i{)}^2-1})=\frac{\pi }{4}$$

带入数据，解得

$$n\approx 1.5002$$