\(\def\bm{\boldsymbol}\)
\(\def\di{\displaystyle}\)
\(\def\ve{\varepsilon_0}\)
\(\def\dd#1#2{\dfrac{\partial #1}{\partial #2}}\)
\(\def\dda#1#2{\dfrac{\partial^2 #1}{\partial #2}}\)
光の速度で、ベクトル解析の公式を使ったので、まとめておきます。
１．はじめに
(1) ベクトルの内積
\(\quad\bm{a}\cdot\bm{b}=a_xb_x+a_yb_y+a_zb_z=\di\sum^3_ia_ib_i\)
(2) ベクトルの外積
\(\quad\bm{a}\times\bm{b}=(a_yb_z-a_zb_y,a_zb_x-a_xb_z,a_xb_y-a_yb_x)\)
\(\quad(\bm{a}\times\bm{b})_i=\di\sum_{j,k=1}^3\epsilon_{i,j,k}\,a_jb_k\)
\begin{align}\hspace{-20mm}\epsilon_{i,j,k}=1&\,:(i,j,k)=(1,2,3),(2,3,1),(3,1,2)\\
-1&\,:(i,j,k)=(1,3,2),(2,1,3),(3,2,1)\\
0&\,: other\end{align}
3. スカラー三重積
\(\quad\bm{A}\cdot(\bm{B}\times\bm{C})=\bm{B}\cdot(\bm{C}\times\bm{A})=\bm{C}\cdot(\bm{A}\times\bm{B})\)
\begin{align}
\bm{A}\cdot(\bm{B}\times\bm{C})&=A_x(\bm{B}\times\bm{C})_x+A_y(\bm{B}\times\bm{C})_y+A_z(\bm{B}\times\bm{C})_z\\[2pt]
&=A_x(B_yC_z-B_zC_y)+A_y(B_zC_x-B_xC_z)+A_z(B_xC_y-B_yC_x)\\[2pt]
&=A_xB_yC_z-A_xB_zC_y+A_yB_zC_x-A_yB_xC_z+A_zB_xC_y-A_zB_yC_x\\[2pt]
&=B_x(C_yA_z-C_zA_y)+B_y(C_zA_x-C_xA_z)+B_z(C_xA_y-C_yA_x)\\[2pt]
&=B_x(\bm{C}\times\bm{A})_x+B_y(\bm{C}\times\bm{A})_y+B_z(\bm{C}\times\bm{A})_z\\
&=\bm{B}\cdot(\bm{C}\times\bm{A})\\[-40pt]
\end{align}
\(\qquad\)また、\(\quad\bm{A}\cdot(\bm{B}\times\bm{C})=\left|\begin{array}{ccc}A_x & A_y & A_z\\ B_x & B_y & B_z\\C_x & C_y & C_z\end{array}\right|\qquad\)である。
4. ベクトル三重積
\(\quad\bm{A}\times(\bm{B}\times\bm{C})=(\bm{A}\cdot\bm{C})\bm{B}-(\bm{A}\cdot\bm{B})\bm{C}\)
\begin{align}
(\bm{A}\times(\bm{B}\times\bm{C}))_x&=A_y(\bm{B}\times\bm{C})_z-A_z(\bm{B}\times\bm{C})_y\\
&=A_y(B_xC_y-B_yC_x)-A_z(B_zC_x-B_xC_z)\\
&=(A_yC_y+A_zB_z)B_x-(A_yB_y+A_zB_z)C_x\\
&=(A_xC_x+A_yC_y+A_zB_z)B_x-(A_xB_x+A_yB_y+A_zB_z)C_x\\
&=(\bm{A}\cdot\bm{C})B_x-(\bm{A}\cdot\bm{B})C_x\\[6pt]
&\hspace{-12mm}\text{同様にして}\\[2pt]
(\bm{A}\times(\bm{B}\times\bm{C}))_y&=(\bm{A}\cdot\bm{C})B_y-(\bm{A}\cdot\bm{B})C_y\\[2pt]
(\bm{A}\times(\bm{B}\times\bm{C}))_z&=(\bm{A}\cdot\bm{C})B_z-(\bm{A}\cdot\bm{B})C_z\\[8pt]
&\hspace{-18mm}\therefore\quad \bm{A}\times(\bm{B}\times\bm{C})=(\bm{A}\cdot\bm{C})\bm{B}-(\bm{A}\cdot\bm{B})\bm{C}
\end{align}
5. 微分演算子
\(\quad\nabla=\left(\,\dd{}{x}\,,\,\dd{}{y}\,,\,\dd{}{z}\,\right)\)
２．勾配、発散、回転
勾配（ grad ）：grad\(\,\phi(\bm{r})=\nabla\,\phi(\bm{r})\equiv\left(\,\dd{}{x}f\,,\,\dd{}{y}f\,,\,\dd{}{z}f\,\right)\)
発散 （ div ）：div\(\,\bm{A}(\bm{r})=\nabla\cdot\bm{A}(\bm{r})\equiv\,\dd{}{x}A_x+\dd{}{y}A_y+\dd{}{z}A_z\)
回転 （ rot ）：rot\(\,\bm{A}(\bm{r})=\left(\dd{}{y}A_z-\dd{}{z}A_y\,,\,\dd{}{z}A_x-\dd{}{x}A_z\,,\,\dd{}{x}A_y-\dd{}{y}A_x\,\right)\)
\(\hspace{8mm}\)または \(\quad(\nabla\times\bm{A})_i=\di\sum_{j,k=1}^3\epsilon_{ijk}(\nabla)_j(\bm{A})_k=\di\sum_{j,k=1}^3\epsilon_{ijk}\,\partial_jA_k\)
３．計算例－１
\(\quad\bm{r}=(x,y,z)\quad r=|\bm{r}|\ne 0\quad\)とする。
(1)\(\,\nabla\,r\)
\begin{align}
\nabla\,r&=\nabla(\sqrt{x^2+y^2+z^2})\\[2pt]
&=\left(\dfrac{x}{\sqrt{x^2+y^2+z^2}}\,,\,\dfrac{y}{\sqrt{x^2+y^2+z^2}}\,,\,\dfrac{z}{\sqrt{x^2+y^2+z^2}}\,\right)\\[2pt]
&=\dfrac{\bm{r}}{r}
\end{align}
(2)\(\,\nabla\left(\dfrac{1}{r}\right)\)
\begin{align}
\nabla\left(\dfrac{1}{r}\right)&=\left(\dd{}{x}\dfrac{1}{\sqrt{x^2+y^2+z^2}}\,,\,\dd{}{y}\dfrac{1}{\sqrt{x^2+y^2+z^2}}\,,\,\dd{}{z}\dfrac{1}{\sqrt{x^2+y^2+z^2}}\,\right)\\[3pt]
&=\left(\,\dfrac{-x}{(x^2+y^2+z^2)^{3/2}}\,,\,\dfrac{-y}{(x^2+y^2+z^2)^{3/2}}\,,\,\dfrac{-z}{(x^2+y^2+z^2)^{3/2}}\,\right)\\[3pt]
&=-\dfrac{\bm{r}}{r^3}
\end{align}
(3)\(\,\nabla\cdot\bm{r}\)
\begin{equation}
\nabla\cdot\bm{r}=\dd{}{x}x+\dd{}{y}y+\dd{}{z}z=3
\end{equation}
(4)\(\,(\nabla\cdot\nabla)\,r\)
\begin{align}
(\nabla\cdot\nabla)r&=\left(\dda{}{x^2}+\dda{}{y^2}+\dda{}{z^2}\right)\sqrt{x^2+y^2+z^2}\\[2pt]
&=\dd{}{x}\dfrac{x}{\sqrt{x^2+y^2+z^2}}+\dd{}{y}\dfrac{y}{\sqrt{x^2+y^2+z^2}}+\dd{}{z}\dfrac{z}{\sqrt{x^2+y^2+z^2}}\\[2pt]
&=\dfrac{1}{\sqrt{x^2+y^2+z^2}}-\dfrac{x^2}{(x^2+y^2+z^2)^{3/2}}\\
& +\dfrac{1}{\sqrt{x^2+y^2+z^2}}-\dfrac{y^2}{(x^2+y^2+z^2)^{3/2}}\\
& +\dfrac{1}{\sqrt{x^2+y^2+z^2}}-\dfrac{z^2}{(x^2+y^2+z^2)^{3/2}}\\[2pt]
&=\dfrac{3}{r}-\dfrac{(x^2+y^2+z^2)}{r^3}=\dfrac{2}{r}
\end{align}
(5)\(\,\nabla\times\bm{r}\)
\begin{equation}
\nabla\times\bm{r}=\left(\dd{}{z}y-\dd{}{y}z\,,\,\dd{}{x}z-\dd{}{z}x\,,\,\dd{}{y}x-\dd{}{x}y\right)=\bm{0}
\end{equation}
(6)\(\,\nabla\times(\nabla\phi)\hspace{10mm}\mathrm{rot}\,\,\mathrm{grad}\,\,\phi\)
\begin{align}
\nabla\times(\nabla\phi)&=\left(\,\dd{}{y}(\dd{\phi}{z})-\dd{}{z}(\dd{\phi}{y})\,\,,\right.\\
&\hspace{20mm}\dd{}{z}(\dd{\phi}{x})-\dd{}{x}(\dd{\phi}{z})\,\,,\\
&\hspace{30mm}\left.\dd{}{x}(\dd{\phi}{y})-\dd{}{y}(\dd{\phi}{x})\,\right)\\
&=0\hspace{40mm}\because\,\,\dd{}{y}(\dd{\phi}{z})=\dd{}{z}(\dd{\phi}{y})
\end{align}
(7)\(\,\nabla\cdot(\nabla\times\bm{A})\hspace{10mm}\mathrm{div}\,\,\mathrm{rot}\,\,\bm{A}\)
\begin{align}
\nabla\cdot(\nabla\times\bm{A})&=\dd{}{x}(\dd{A_z}{y}-\dd{A_y}{z})+\dd{}{y}(\dd{A_x}{z}-\dd{A_z}{x})+\dd{}{z}(\dd{A_y}{x}-\dd{A_x}{y})\\[3pt]
&=\dfrac{\partial^2A_z}{\partial x\partial y}-\dfrac{\partial^2A_y}{\partial x\partial z}+\dfrac{\partial^2A_x}{\partial y\partial z}-\dfrac{\partial^2A_z}{\partial y\partial x}+\dfrac{\partial^2A_y}{\partial z\partial x}-\dfrac{\partial^2A_x}{\partial z\partial y}\\[3pt]
&=0\hspace{40mm}\because\,\,\,\dfrac{\partial^2A_z}{\partial x\partial y}=\dfrac{\partial^2A_z}{\partial y\partial x}
\end{align}
(8)\(\,\nabla\times(\nabla\times\bm{A})\hspace{10mm}\mathrm{rot}\,\,\mathrm{rot}\,\,\bm{A}\)
\begin{align}
(\nabla\times(\nabla\times\bm{A}))_x&=\dd{}{y}\left(\dd{A_y}{x}-\dd{A_x}{y}\right)-\dd{}{z}\left(\dd{A_x}{z}-\dd{A_z}{x}\right)\\[3pt]
&=\dfrac{\partial^2}{\partial y\partial x}A_y-\dda{}{y^2}A_x-\dda{}{z^2}A_x+\dfrac{\partial^2}{\partial z\partial x}A_z\\[3pt]
&=\dda{}{x^2}A_x+\dfrac{\partial^2}{\partial y\partial x}A_y+\dfrac{\partial^2}{\partial z\partial x}A_z-\left(\dda{}{x^2}+\dda{}{y^2}+\dda{}{z^2}\right)A_x\\[3pt]
&=\dd{}{x}\left(\dd{A_x}{x}+\dd{A_y}{y}+\dd{A_z}{z}\right)-\Delta A_x\\[3pt]
&=\dd{}{x}(\nabla\cdot\bm{A})-\Delta A_x\\[3pt]
&\hspace{-20mm}\text{同様に}\\
(\nabla\times(\nabla\times\bm{A}))_y&=\dd{}{y}(\nabla\cdot\bm{A})-\Delta A_y\\[3pt]
(\nabla\times(\nabla\times\bm{A}))_z&=\dd{}{z}(\nabla\cdot\bm{A})-\Delta A_z\\[3pt]
&\hspace{-20mm}\text{よって}\\[2pt]
\nabla\times(\nabla\times\bm{A})&=\nabla(\nabla\cdot\bm{A})-\Delta\bm{A}
\end{align}