様々な微分演算子に関する公式｜大学・理系（大学数学/大学物理）【レポート代行・課題代行・宿題代行・家庭教師】

次の関係式が成り立つことを示せ。
$\begin{aligned} d i v (f (\vec{r}) \vec{A} (\vec{r})) & = f (\vec{r}) d i v \vec{A} (\vec{r}) + g r a d f (\vec{r}) \cdot \vec{A} (\vec{r}) \\ r o t r o t \vec{A} (\vec{r}) & = g r a d d i v \vec{A} (\vec{r}) - Δ \vec{A} (\vec{r}) \\ d i v (\vec{A} (\vec{r}) \times \vec{B} (\vec{r})) & = \vec{B} (\vec{r}) \cdot r o t \vec{A} (\vec{r}) - \vec{A} (\vec{r}) \cdot r o t \vec{B} (\vec{r}) \\ r o t g r a d f (\vec{r}) & = 0 \\ d i v r o t \vec{A} (\vec{r}) & = 0 \end{aligned}$

Einstein の縮約、及び Levi-Civita の記号を利用して証明を行う。（こちらの記事を参照）

$\begin{aligned} d i v (f (\vec{r}) \vec{A} (\vec{r})) & = \frac{\partial}{\partial x_{i}} (f (\vec{r}) A_{x_{i}} (\vec{r})) \\ = \frac{\partial f (\vec{r})}{\partial x_{i}} A_{x_{i}} (\vec{r}) + f (\vec{r}) \frac{\partial A_{x_{i}} (\vec{r})}{\partial x_{i}} \\ = g r a d f (\vec{r}) \cdot \vec{A} (\vec{r}) + f (\vec{r}) d i v \vec{A} (\vec{r}) \end{aligned}$

$\begin{aligned} {(r o t r o t \vec{A} (\vec{r}))}_{x_{i}} & = ϵ_{i j k} \frac{\partial}{\partial x_{j}} {(r o t \vec{A} (\vec{r}))}_{x_{k}} \\ = ϵ_{i j k} \frac{\partial}{\partial x_{j}} ϵ_{k l m} \frac{\partial A_{x_{m}} (\vec{r})}{\partial x_{l}} \\ = (ϵ_{i j k} ϵ_{l m k}) \frac{\partial}{\partial x_{j}} \frac{\partial A_{x_{m}} (\vec{r})}{\partial x_{l}} \\ = (δ_{i, l} δ_{j, m} - δ_{i, m} δ_{j, l}) \frac{\partial^{2} A_{x_{m}} (\vec{r})}{\partial x_{j} \partial x_{l}} \\ = \frac{\partial^{2} A_{x_{j}} (\vec{r})}{\partial x_{j} \partial x_{i}} - \frac{\partial^{2} A_{x_{i}} (\vec{r})}{\partial x_{j} \partial x_{j}} \\ = {(g r a d d i v \vec{A} (\vec{r}))}_{x_{i}} - Δ {\vec{A}}_{x_{i}} (\vec{r}) \end{aligned}$

$\begin{aligned} d i v (\vec{A} (\vec{r}) \times \vec{B} (\vec{r})) & = \frac{\partial}{\partial x_{i}} {(\vec{A} (\vec{r}) \times \vec{B} (\vec{r}))}_{x_{i}} \\ = \frac{\partial}{\partial x_{i}} (ϵ_{i j k} A_{x_{j}} (\vec{r}) B_{x_{k}} (\vec{r})) \\ = ϵ_{k i j} \frac{\partial A_{x_{j}} (\vec{r})}{\partial x_{i}} B_{x_{k}} (\vec{r}) - A_{x_{j}} (\vec{r}) ϵ_{j i k} \frac{\partial B_{x_{k}} (\vec{r})}{\partial x_{i}} \\ = r o t \vec{A} (\vec{r}) \cdot \vec{B} (\vec{r}) - \vec{A} \cdot r o t \vec{B} (\vec{r}) \end{aligned}$

$\begin{aligned} {(r o t g r a d f (\vec{r}))}_{x_{i}} & = ϵ_{i j k} \frac{\partial}{\partial x_{j}} {(g r a d f (\vec{r}))}_{x_{k}} \\ = ϵ_{i j k} \frac{\partial^{2}}{\partial x_{j} \partial x_{k}} f (\vec{r}) \\ = ϵ_{i k j} \frac{\partial^{2}}{\partial x_{k} \partial x_{j}} f (\vec{r}) \\ = - ϵ_{i j k} \frac{\partial^{2}}{\partial x_{j} \partial x_{k}} f (\vec{r}) \\ = 0 \end{aligned}$
ここで2行目から3行目において $j$ と $k$ を入れ替えても値が変わらないことを用いた。

$\begin{aligned} d i v r o t \vec{A} (\vec{r}) & = \frac{\partial}{\partial x_{i}} {(r o t \vec{A} (\vec{r}))}_{x_{i}} \\ = \frac{\partial}{\partial x_{i}} ϵ_{i j k} \frac{\partial A_{x_{k}} (\vec{r})}{\partial x_{j}} \\ = ϵ_{i j k} \frac{\partial^{2} A_{x_{k}} (\vec{r})}{\partial x_{i} \partial x_{j}} \\ = 0 \end{aligned}$
ここで3行目から4行目は $i$ と $j$ を入れ替えても値は変わらないことを用いた。