向量三重积$a\times(b\times c)$

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定义Kronecker delta$$\delta_{ij}=\begin{cases}1&i=j\\0&i\ne j\end{cases}$$
Levi-Civita symbol$$ϵ_{ijk}=\begin{cases}1&(ijk)为(123)的偶置换\\-1&(ijk)为(123)的奇置换\\0&i,j,k有两个相等\end{cases}$$
定义向量外积$$\boldsymbol{a} \times \boldsymbol{b}=\operatorname{det}\left|\begin{array}{ccc}\hat{\boldsymbol{e}}_{1} & \hat{\boldsymbol{e}}_{2} & \hat{\boldsymbol{e}}_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|=\epsilon_{i j k} \hat{\boldsymbol{e}}_{i} a_{j} b_{k}$$

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两个Levi-Civita symbol乘积可表示为Kronecker delta\begin{align}\label1 \epsilon_{i j k} \epsilon_{l m n}= & +\delta_{i l} \delta_{j m} \delta_{k n}+\delta_{i m} \delta_{j n} \delta_{k l}+\delta_{i n} \delta_{j l} \delta_{k m} \\ \nonumber& -\delta_{i m} \delta_{j l} \delta_{k n}-\delta_{i l} \delta_{j n} \delta_{k m}-\delta_{i n} \delta_{j m} \delta_{k l}\end{align}令$i=l$得$ϵ_{ijk}ϵ_{imn}=δ_{jm}δ_{kn}-δ_{jn}δ_{km}$

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证明\begin{aligned} \epsilon_{i j k} \epsilon_{i m n} & =\delta_{i i}\left(\delta_{j m} \delta_{k n}-\delta_{j n} \delta_{k m}\right)+\delta_{i m} \delta_{j n} \delta_{k i}+\delta_{i n} \delta_{j i} \delta_{k m}-\delta_{i m} \delta_{j i} \delta_{k n}-\delta_{i n} \delta_{j m} \delta_{k i} \\ & =3\left(\delta_{j m} \delta_{k n}-\delta_{j n} \delta_{k m}\right)+\delta_{k m} \delta_{j n}+\delta_{j n} \delta_{k m}-\delta_{j m} \delta_{k n}-\delta_{k n} \delta_{j m} \\ & =\delta_{j m} \delta_{k n}-\delta_{j n} \delta_{k m}\end{aligned}令$i = l ,j = m$得$\epsilon_{i j k} \epsilon_{i j n}=2 \delta_{k n}$
令$i=l,j=m,k=n$得$\epsilon_{i j k} \epsilon_{i j k}=6$

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用\eqref{1}可以证明$a \times(b \times c)=b(a \cdot c)-c(a \cdot b)$如下
\begin{aligned} \boldsymbol{d} & =\boldsymbol{a} \times(\boldsymbol{b} \times \boldsymbol{c}) \\ d_{m} & =\epsilon_{m n i} a_{n}\left(\epsilon_{i j k} b_{j} c_{k}\right) \\ & =\epsilon_{i m n} \epsilon_{i j k} a_{n} b_{j} c_{k} \\ & =\left(\delta_{m j} \delta_{n k}-\delta_{m k} \delta_{n j}\right) a_{n} b_{j} c_{k} \\ & =b_{m} a_{k} c_{k}-c_{m} a_{j} b_{j} \\ & =[\boldsymbol{b}(\boldsymbol{a} \cdot \boldsymbol{c})]_{m}-[\boldsymbol{c}(\boldsymbol{a} \cdot \boldsymbol{b})]_{m}\end{aligned}同样地\begin{aligned} {[\boldsymbol{\nabla} \times(\boldsymbol{\nabla} \times \boldsymbol{a})]_{i} } & =\epsilon_{i j k} \partial_{j} \epsilon_{k m n} \partial_{m} a_{n} \\ & =\epsilon_{k i j} \epsilon_{k m n} \partial_{j} \partial_{m} a_{n} \\ & =\partial_{j} \partial_{i} a_{j}-\partial_{j} \partial_{j} a_{i} \\ & =\partial_{i} \partial_{j} a_{j}-\partial_{j} \partial_{j} a_{i} \\ & =\left[\nabla(\nabla \cdot \boldsymbol{a})-\nabla^{2} \boldsymbol{a}\right]_{i}\end{aligned}

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原文(9)式$\epsilon_{i j k} \epsilon_{i j k}=6$这个错了吧