电动力学的数学基础之一是矢量分析。本文介绍推导矢量分析公式的另一种方法——指标法。指标法的本质是通过爱因斯坦求和约定，将各个坐标分量简记为一通式，并省略求和符号。指标法充分利用各分量的对称性，因而形式上较为简单，运算上难以出错，且能够处理较复杂的算式和高阶张量。由于完全在笛卡尔坐标系下进行运算，本文不考虑指标的协变与逆变。
求和约定：当式子中任何一个指标出现了两次，并且一次为上标，一次为下标，那么该式表示的实际上是对这个角标一切可能值的求和。
哑指标：在表达式的某一单项式之中出现且仅出现 2 2 2 哑指标 。哑指标的符号与表达式的值无关。对这种指标应当进行遍历求和。哑指标可以把多项式缩写成一项。
自由指标：在表达式或方程中自由指标 可以出现多次，但不得在同项内重复出现两次。自由指标可以把多个方程缩写成一个方程。
指标取值：一般拉丁字母 i , j , k , l , m , n , p , q , r i,j,k,l,m,n,p,q,r i , j , k , l , m , n , p , q , r 1 , 2 , 3 1,2,3 1 , 2 , 3 μ , ν , ρ , σ , τ \mu,\nu,\rho,\sigma,\tau μ , ν , ρ , σ , τ 0 , 1 , 2 , 3 0,1,2,3 0 , 1 , 2 , 3 1 , 2 , 3 , 4 1,2,3,4 1 , 2 , 3 , 4
向量的表示
向量 α = a 1 e 1 + a 2 e 2 + a 3 e 3 \boldsymbol{\alpha} = a_1 \bm{e_1} + a_2 \bm{e_2} + a_3 \bm{e_3} α = a 1 e 1 + a 2 e 2 + a 3 e 3
α = a i e i \boldsymbol{\alpha} = a_i \boldsymbol{e}_i
α = a i e i
这便代表了
α = ∑ i = 1 3 a i e i = ∑ j = 1 3 a j e j = ∑ k = 1 3 a k e k \boldsymbol{\alpha} = \sum_{i=1}^3 a_i \boldsymbol{e}_i = \sum_{j=1}^3 a_j \boldsymbol{e}_j = \sum_{k=1}^3 a_k \boldsymbol{e}_k
α = i = 1 ∑ 3 a i e i = j = 1 ∑ 3 a j e j = k = 1 ∑ 3 a k e k
i i i j , k j,k j , k i i i
方程的表示
一个三元一次线性方程组
{ A 11 x 1 + A 12 x 2 + A 13 x 3 = b 1 A 21 x 1 + A 22 x 2 + A 23 x 3 = b 2 A 31 x 1 + A 32 x 2 + A 33 x 3 = b 3 \left\{\begin{matrix}
A_{11} x_{1}+A_{12} x_{2}+A_{13} x_{3}=b_{1} \\
A_{21} x_{1}+A_{22} x_{2}+A_{23} x_{3}=b_{2} \\
A_{31} x_{1}+A_{32} x_{2}+A_{33} x_{3}=b_{3}
\end{matrix}\right.
⎩ ⎪ ⎨ ⎪ ⎧ A 1 1 x 1 + A 1 2 x 2 + A 1 3 x 3 = b 1 A 2 1 x 1 + A 2 2 x 2 + A 2 3 x 3 = b 2 A 3 1 x 1 + A 3 2 x 2 + A 3 3 x 3 = b 3
可以利用求和约定简记为
A i j x j = b i A_{ij} x_j =b_i
A i j x j = b i
其中 j j j i i i
Kronecker Delta 在线性代数里常用于表现正交性，定义为
δ i j = { 1 if i = j 0 if i ≠ j \delta_{ij}=\begin{cases}
1 &\text{if} \;i=j \\
0 &\text{if} \;i \neq j
\end{cases}
δ i j = { 1 0 if i = j if i = j
δ i m δ m j = δ i j \delta_{im} \delta_{mj} = \delta_{ij} δ i m δ m j = δ i j
Proof
按照爱因斯坦求和约定将其展开，得到
δ i m δ m j = δ i 1 δ 1 j + δ i 2 δ 2 j + δ 31 δ 3 j \delta_{im} \delta_{mj} = \delta_{i1}\delta_{1j} + \delta_{i2}\delta_{2j} + \delta_{31}\delta_{3j}
δ i m δ m j = δ i 1 δ 1 j + δ i 2 δ 2 j + δ 3 1 δ 3 j
由 Kronecker Delta 的定义，对于第一项，若 i = j = 1 i=j=1 i = j = 1 δ i 1 δ 1 j = 1 \delta_{i1}\delta_{1j}=1 δ i 1 δ 1 j = 1 0 0 0 1 1 1
否则 i = j = 2 , 3 i=j=2,3 i = j = 2 , 3 i ≠ j i \neq j i = j i = j = 1 i=j=1 i = j = 1 0 0 0 δ i j \delta_{ij} δ i j δ i m δ m j = δ i j \delta_{im} \delta_{mj} = \delta_{ij} δ i m δ m j = δ i j
ε i j k δ k m = ε i j m \varepsilon^{ijk} \delta_{km} = \varepsilon^{ijm} ε i j k δ k m = ε i j m
Proof
按照爱因斯坦求和约定将其展开，得到
ε i j k δ k m = ε i j 1 δ 1 m + ε i j 2 δ 2 m + ε i j 3 δ 3 m \varepsilon^{ijk} \delta_{km} = \varepsilon^{ij1} \delta_{1m}+ \varepsilon^{ij2} \delta_{2m} + \varepsilon^{ij3} \delta_{3m}
ε i j k δ k m = ε i j 1 δ 1 m + ε i j 2 δ 2 m + ε i j 3 δ 3 m
由此可见，m = 1 , 2 , 3 m=1,2,3 m = 1 , 2 , 3 0 0 0 R H S RHS R H S
假如 O x y z Oxyz O x y z e 1 , e 2 , e 3 \bm{e}_1,\bm{e}_2,\bm{e}_3 e 1 , e 2 , e 3
e 1 = [ 1 0 0 ] e_1=\begin{bmatrix}1 &0 & 0\end{bmatrix}
e 1 = [ 1 0 0 ]
e 2 = [ 0 1 0 ] e_2 =\begin{bmatrix}0 &1 & 0\end{bmatrix}
e 2 = [ 0 1 0 ]
e 3 = [ 0 0 1 ] e_3 =\begin{bmatrix}0 &0 & 1\end{bmatrix}
e 3 = [ 0 0 1 ]
也就是
e i = [ δ i 1 δ i 2 δ i 3 ] e_i = \begin{bmatrix} \delta_{i1} & \delta_{i2} & \delta_{i3} \end{bmatrix}
e i = [ δ i 1 δ i 2 δ i 3 ]
定义符号 ε i j k \varepsilon_{ijk} ε i j k
ε i j k = { 1 if ( i , j , k ) is an even permutation of ( 1 , 2 , 3 ) − 1 if ( i , j , k ) is an odd permutation of ( 1 , 2 , 3 ) 0 if two or more indices are equal \varepsilon_{ijk}=\begin{cases}
1 &\text{if} \; (i,j,k) \; \text{is an even permutation of} \; (1,2,3) \\
-1 &\text{if} \; (i,j,k) \; \text{is an odd permutation of} \; (1,2,3) \\
0 &\text{if two or more indices are equal}
\end{cases}
ε i j k = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 1 − 1 0 if ( i , j , k ) is an even permutation of ( 1 , 2 , 3 ) if ( i , j , k ) is an odd permutation of ( 1 , 2 , 3 ) if two or more indices are equal
显然有
ε i j k ε l m n = { 1 if ( i , j , k ) and ( l , m , n ) have the same parity − 1 if ( i , j , k ) and ( l , m , n ) have the opposite parity 0 if two or more indices in one symbol are equal \varepsilon_{i j k} \varepsilon_{l m n}=
\begin{cases}
1 &\text{if} \; (i,j,k) \; \text{and} \; (l,m,n) \; \text{have the same parity} \\
-1 &\text{if} \; (i,j,k) \; \text{and} \; (l,m,n) \; \text{have the opposite parity} \\
0 &\text{if two or more indices in one symbol are equal}
\end{cases}
ε i j k ε l m n = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ 1 − 1 0 if ( i , j , k ) and ( l , m , n ) have the same parity if ( i , j , k ) and ( l , m , n ) have the opposite parity if two or more indices in one symbol are equal
还可以推出公式
ε i j k ε l m n = ∣ δ i l δ i m δ i n δ j l δ j m δ j n δ k l δ k m δ k n ∣ \varepsilon_{i j k} \varepsilon_{l m n}=
\begin{vmatrix}
\delta_{i l} & \delta_{i m} & \delta_{i n} \\
\delta_{j l} & \delta_{j m} & \delta_{j n} \\
\delta_{k l} & \delta_{k m} & \delta_{k n}
\end{vmatrix}
ε i j k ε l m n = ∣ ∣ ∣ ∣ ∣ ∣ ∣ δ i l δ j l δ k l δ i m δ j m δ k m δ i n δ j n δ k n ∣ ∣ ∣ ∣ ∣ ∣ ∣
其中 δ i j \delta_{ij} δ i j
Proof
公式的证明最好等到阅读了行列式 后。
注意到 ε i j k \varepsilon_{i j k} ε i j k e i ⋅ ( e j × e k ) \bm{e}_i \cdot (\bm{e}_j \times \bm{e}_k) e i ⋅ ( e j × e k )
ε i j k = e i ⋅ ( e j × e k ) = ∣ δ 1 i δ 2 i δ 3 i δ 1 j δ 2 j δ 3 j δ 1 k δ 2 k δ 3 k ∣ \varepsilon_{i j k} = \bm{e}_i \cdot (\bm{e}_j \times \bm{e}_k) =
\begin{vmatrix}
\delta^i_1 & \delta^i_2 & \delta^i_3 \\
\delta^j_1 & \delta^j_2 & \delta^j_3 \\
\delta^k_1 & \delta^k_2 & \delta^k_3
\end{vmatrix}
ε i j k = e i ⋅ ( e j × e k ) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ δ 1 i δ 1 j δ 1 k δ 2 i δ 2 j δ 2 k δ 3 i δ 3 j δ 3 k ∣ ∣ ∣ ∣ ∣ ∣ ∣
ε l m n = e l ⋅ ( e m × e n ) = ∣ δ 1 l δ 2 l δ 3 l δ 1 m δ 2 m δ 3 m δ 1 n δ 2 n δ 3 n ∣ = ∣ δ 1 l δ 1 m δ 1 n δ 2 l δ 2 m δ 2 n δ 3 l δ 3 m δ 3 n ∣ \varepsilon_{l m n} = \bm{e}_l \cdot (\bm{e}_m \times \bm{e}_n) =
\begin{vmatrix}
\delta^l_1 & \delta^l_2 & \delta^l_3 \\
\delta^m_1 & \delta^m_2 & \delta^m_3 \\
\delta^n_1 & \delta^n_2 & \delta^n_3
\end{vmatrix}
=
\begin{vmatrix}
\delta^l_1 & \delta^m_1 & \delta^n_1 \\
\delta^l_2 & \delta^m_2 & \delta^n_2 \\
\delta^l_3 & \delta^m_3 & \delta^n_3
\end{vmatrix}
ε l m n = e l ⋅ ( e m × e n ) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ δ 1 l δ 1 m δ 1 n δ 2 l δ 2 m δ 2 n δ 3 l δ 3 m δ 3 n ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ δ 1 l δ 2 l δ 3 l δ 1 m δ 2 m δ 3 m δ 1 n δ 2 n δ 3 n ∣ ∣ ∣ ∣ ∣ ∣ ∣
由线性代数，有 ∣ A ∣ ∣ B ∣ = ∣ A B ∣ |\bm{A}||\bm{B}|=|\bm{AB}| ∣ A ∣ ∣ B ∣ = ∣ A B ∣
ε i j k ε l m n = ∣ δ 1 i δ 2 i δ 3 i δ 1 j δ 2 j δ 3 j δ 1 k δ 2 k δ 3 k ∣ ∣ δ 1 l δ 1 m δ 1 n δ 2 l δ 2 m δ 2 n δ 3 l δ 3 m δ 3 n ∣ = ∣ δ r i δ l r δ r i δ m r δ r i δ n r δ r j δ l r δ r j δ m r δ r j δ n r δ r k δ l r δ r k δ m r δ r k δ n r ∣ \varepsilon_{i j k} \varepsilon_{l m n} = \begin{vmatrix}
\delta^i_1 & \delta^i_2 & \delta^i_3 \\
\delta^j_1 & \delta^j_2 & \delta^j_3 \\
\delta^k_1 & \delta^k_2 & \delta^k_3
\end{vmatrix}
\begin{vmatrix}
\delta^l_1 & \delta^m_1 & \delta^n_1 \\
\delta^l_2 & \delta^m_2 & \delta^n_2 \\
\delta^l_3 & \delta^m_3 & \delta^n_3
\end{vmatrix}
=
\begin{vmatrix}
\delta^i_r \delta^r_l & \delta^i_r \delta^r_m & \delta^i_r \delta^r_n \\
\delta^j_r \delta^r_l & \delta^j_r \delta^r_m & \delta^j_r\delta^r_n \\
\delta^k_r \delta^r_l & \delta^k_r \delta^r_m & \delta^k_r \delta^r_n
\end{vmatrix}
ε i j k ε l m n = ∣ ∣ ∣ ∣ ∣ ∣ ∣ δ 1 i δ 1 j δ 1 k δ 2 i δ 2 j δ 2 k δ 3 i δ 3 j δ 3 k ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ δ 1 l δ 2 l δ 3 l δ 1 m δ 2 m δ 3 m δ 1 n δ 2 n δ 3 n ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ δ r i δ l r δ r j δ l r δ r k δ l r δ r i δ m r δ r j δ m r δ r k δ m r δ r i δ n r δ r j δ n r δ r k δ n r ∣ ∣ ∣ ∣ ∣ ∣ ∣
其中每个 r r r 1 1 1
ε i j k ε l m n = ∣ δ l i δ m i δ n i δ l j δ m j δ n j δ l k δ m k δ n k ∣ \varepsilon_{i j k} \varepsilon_{l m n}=
\begin{vmatrix}
\delta^i_l & \delta^i_m & \delta^i_n \\
\delta^j_l & \delta^j_m & \delta^j_n \\
\delta^k_l & \delta^k_m & \delta^k_n
\end{vmatrix}
ε i j k ε l m n = ∣ ∣ ∣ ∣ ∣ ∣ ∣ δ l i δ l j δ l k δ m i δ m j δ m k δ n i δ n j δ n k ∣ ∣ ∣ ∣ ∣ ∣ ∣
ε i j k ε i m n = δ j m δ k n − δ j n δ k m \varepsilon_{i j k} \varepsilon_{i m n}=\delta_{j m} \delta_{k n}-\delta_{j n} \delta_{k m} ε i j k ε i m n = δ j m δ k n − δ j n δ k m ε i j k ε p q k = δ i p δ j q − δ i q δ j p \varepsilon_{i j k} \varepsilon_{pqk}=\delta_{i p} \delta_{j q}-\delta_{i q} \delta_{j p} ε i j k ε p q k = δ i p δ j q − δ i q δ j p
对于向量来说，逆变向量 contravariant vector 应当写上标，协变向量 covariant vector 应当写下标。
对于矩阵来说，矩阵 M M M i i i j j j M j i M^i_j M j i
v i = [ v 1 v 2 ⋮ v n ] , v j = [ v 1 v 2 ⋯ v n ] \bm{v}^i= \begin{bmatrix}
v^1 \\
v^2 \\
\vdots \\
v^n
\end{bmatrix}
, \;
\bm{v}_j = \begin{bmatrix}
v_1 & v_2 & \cdots & v_n
\end{bmatrix}
v i = ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ v 1 v 2 ⋮ v n ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ , v j = [ v 1 v 2 ⋯ v n ]
行列式的逆序数定义为
det M = ∣ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋱ ⋮ a n 1 a n 2 ⋯ a n n ∣ = ∑ j 1 j 2 ⋯ j n ( − 1 ) τ ( j 1 j 2 ⋯ j n ) a 1 j 1 a 2 j 2 ⋯ a n j n \det M=
\begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1 n} \\
a_{21} & a_{22} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{vmatrix}=
\sum_{j_{1} j_{2} \cdots j_{n}}(-1)^{\tau\left(j_{1} j_{2} \cdots j_{n}\right)} a_{1 j_{1}} a_{2 j_{2}} \cdots a_{n j_{n}}
det M = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ a 1 1 a 2 1 ⋮ a n 1 a 1 2 a 2 2 ⋮ a n 2 ⋯ ⋯ ⋱ ⋯ a 1 n a 2 n ⋮ a n n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = j 1 j 2 ⋯ j n ∑ ( − 1 ) τ ( j 1 j 2 ⋯ j n ) a 1 j 1 a 2 j 2 ⋯ a n j n
每一项前面的系数 ( − 1 ) τ ( j 1 j 2 ⋯ j n ) (-1)^{\tau \left(j_{1} j_{2} \cdots j_{n}\right)} ( − 1 ) τ ( j 1 j 2 ⋯ j n ) ε j 1 j 2 ⋯ j n \varepsilon_{j_1 j_2 \cdots j_n} ε j 1 j 2 ⋯ j n
det M = ε i j k a i 1 a j 2 a k 3 \det M = \varepsilon^{ijk} a^1_i a^2_ja^3_k
det M = ε i j k a i 1 a j 2 a k 3
对于给定行指标的一个排列 i 1 i 2 ⋯ i n i_1 i_2 \cdots i_n i 1 i 2 ⋯ i n
det M = ∑ j 1 j 2 ⋯ j n ( − 1 ) τ ( i 1 i 2 ⋯ i n ) + τ ( j 1 j 2 ⋯ j n ) a i 1 j 1 a i 2 j 2 ⋯ a i n j n \det M=
\sum_{j_{1} j_{2} \cdots j_{n}}(-1)^{\tau\left(i_{1} i_{2} \cdots i_{n}\right)+\tau\left(j_{1} j_{2} \cdots j_{n}\right)} a_{i_1 j_{1}} a_{i_2 j_{2}} \cdots a_{i_n j_{n}}
det M = j 1 j 2 ⋯ j n ∑ ( − 1 ) τ ( i 1 i 2 ⋯ i n ) + τ ( j 1 j 2 ⋯ j n ) a i 1 j 1 a i 2 j 2 ⋯ a i n j n
于是再次利用爱因斯坦求和约定，有
ε p q r det M = ε i j k a i p a j q a k r \varepsilon_{pqr} \det M= \varepsilon^{ijk} a^p_i a^q_j a^r_k
ε p q r det M = ε i j k a i p a j q a k r
注意全指标求和的仅有列元素，因此 ε p q r \varepsilon_{pqr} ε p q r
基矢 e i \bm{e}_{i} e i e j \bm{e}_{j} e j
e i ⋅ e j = δ i j \bm{e}_{i} \cdot \bm{e}_{j}=\delta_{i j}
e i ⋅ e j = δ i j
向量 a \bm{a} a b \bm{b} b
a ⋅ b = a i e i b j e j = a i b j δ i j \bm{a} \cdot \bm{b} = a_i \bm{e}_i b^j \bm{e}^j = a_ib_j\delta_{ij}
a ⋅ b = a i e i b j e j = a i b j δ i j
基矢 e i \bm{e}_{i} e i e j \bm{e}_{j} e j
e i × e j = ∣ e 1 e 2 e 3 δ 1 i δ 2 i δ 3 i δ 1 j δ 2 j δ 3 j ∣ = ε p q r e p δ q i δ r j \bm{e}_{i} \times \bm{e}_{j}=
\begin{vmatrix}
\bm{e}_{1} & \bm{e}_{2} & \bm{e}_{3} \\
\delta^i_1 & \delta^i_2 & \delta^i_3 \\
\delta^j_1 & \delta^j_2 & \delta^j_3
\end{vmatrix}
= \varepsilon^{pqr} \bm{e}_{p} \delta^{i}_{q} \delta^{j}_{r}
e i × e j = ∣ ∣ ∣ ∣ ∣ ∣ ∣ e 1 δ 1 i δ 1 j e 2 δ 2 i δ 2 j e 3 δ 3 i δ 3 j ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ε p q r e p δ q i δ r j
向量 a \bm{a} a b \bm{b} b
a × b = ∣ a 1 a 2 a 3 b 1 b 2 b 3 e 1 e 2 e 3 ∣ = ε p q r a p b q e r \bm{a} \times \bm{b} =
\begin{vmatrix}
a_1 & a_2 &a_3 \\
b_1 & b_2 &b_3 \\
\bm{e}_{1} & \bm{e}_{2} & \bm{e}_{3}
\end{vmatrix}
= \varepsilon^{pqr} a_p b_q \bm{e}_r
a × b = ∣ ∣ ∣ ∣ ∣ ∣ ∣ a 1 b 1 e 1 a 2 b 2 e 2 a 3 b 3 e 3 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ε p q r a p b q e r
基矢 e 1 , e 2 , e 3 \bm{e}_1,\bm{e}_2,\bm{e}_3 e 1 , e 2 , e 3
e i ⋅ ( e j × e k ) = ∣ δ 1 i δ 2 i δ 3 i δ 1 j δ 2 j δ 3 j δ 1 k δ 2 k δ 3 k ∣ = ε p q r δ p i δ q j δ r k \bm{e}_i \cdot (\bm{e}_j \times \bm{e}_k) =
\begin{vmatrix}
\delta^i_1 & \delta^i_2 & \delta^i_3 \\
\delta^j_1 & \delta^j_2 & \delta^j_3 \\
\delta^k_1 & \delta^k_2 & \delta^k_3
\end{vmatrix}
= \varepsilon_{pqr} \delta^i_p \delta^j_q \delta^k_r
e i ⋅ ( e j × e k ) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ δ 1 i δ 1 j δ 1 k δ 2 i δ 2 j δ 2 k δ 3 i δ 3 j δ 3 k ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ε p q r δ p i δ q j δ r k
矢量 a , b , c \bm{a}, \bm{b}, \bm{c} a , b , c
a ⋅ ( b × c ) = ∣ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ∣ = ε p q r a p b q c r \boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c}) =
\begin{vmatrix}
a_1 & a_2 &a_3 \\
b_1 & b_2 &b_3 \\
c_1 & c_2 &c_3
\end{vmatrix}
= \varepsilon_{pqr} a_p b_q c_r
a ⋅ ( b × c ) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ε p q r a p b q c r
Proof
先算叉乘，再算点乘
a ⋅ ( b × c ) = a i ( b × c ) j δ i j = a i ε p q j b p c q δ i j = ε p q i a i b p c q \begin{aligned}
\boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c}) &= a_i(\boldsymbol{b} \times \boldsymbol{c})_j \delta_{ij} \\
&= a_i\varepsilon^{p q j} b_p c_q \delta_{ij} \\
&= \varepsilon^{p q i} a_i b_p c_q
\end{aligned}
a ⋅ ( b × c ) = a i ( b × c ) j δ i j = a i ε p q j b p c q δ i j = ε p q i a i b p c q
我们熟知
a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c \bm{a} \times (\bm{b} \times \bm{c}) = (\bm{a} \cdot \bm{c}) \bm{b} - (\bm{a} \cdot \bm{b}) \bm{c}
a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c
下面证明这个式子
a × ( b × c ) = ε i j k a i ( b × c ) j e k = ε i j k a i ε p q j b p c q e k = ε i j k ε p q j a i b p c q e k = − ε i k j ε p q j a i b p c q e k \begin{aligned}
\bm{a} \times (\bm{b} \times \bm{c}) &= \varepsilon_{ijk}a_i (\boldsymbol{b} \times \boldsymbol{c})_j \bm{e}_k \\
&= \varepsilon_{ijk} a_i\varepsilon_{pqj}b_p c_q \bm{e}_k \\
&= \varepsilon_{ijk}\varepsilon_{pqj} a_i b_p c_q \bm{e}_k \\
&= -\varepsilon_{ikj}\varepsilon_{pqj} a_i b_p c_q \bm{e}_k
\end{aligned}
a × ( b × c ) = ε i j k a i ( b × c ) j e k = ε i j k a i ε p q j b p c q e k = ε i j k ε p q j a i b p c q e k = − ε i k j ε p q j a i b p c q e k
由推论继而得到
a × ( b × c ) = ( δ q i δ p k − δ p i δ q k ) a i b p c q e k = δ q i δ p k a i b p c q e k − δ p i δ q k a i b p c q e k = ( δ q i a i c q ) b p ( δ p k e k ) − ( δ p i a i b p ) c q ( δ q k e k ) = ( a ⋅ c ) b p e p − ( a ⋅ b ) c q e q = ( a ⋅ c ) b − ( a ⋅ b ) c \begin{aligned}
\bm{a} \times (\bm{b} \times \bm{c}) &= (\delta^i_q\delta^k_p-\delta^i_p \delta^k_q)a_i b_p c_q \bm{e}_k\\
&= \delta^i_q\delta^k_p a_i b_p c_q \bm{e}_k-\delta^i_p \delta^k_q a_i b_p c_q \bm{e}_k \\
&= (\delta^i_q a_i c_q) b_p (\delta^k_p \bm{e}_k)- (\delta^i_p a_i b_p) c_q (\delta^k_q \bm{e}_k ) \\
&= (\bm{a} \cdot \bm{c}) b_p\bm{e}_p- (\bm{a} \cdot \bm{b}) c_q \bm{e}_q \\
&= (\bm{a} \cdot \bm{c}) \bm{b} - (\bm{a} \cdot \bm{b}) \bm{c}
\end{aligned}
a × ( b × c ) = ( δ q i δ p k − δ p i δ q k ) a i b p c q e k = δ q i δ p k a i b p c q e k − δ p i δ q k a i b p c q e k = ( δ q i a i c q ) b p ( δ p k e k ) − ( δ p i a i b p ) c q ( δ q k e k ) = ( a ⋅ c ) b p e p − ( a ⋅ b ) c q e q = ( a ⋅ c ) b − ( a ⋅ b ) c
哈密顿算子
∇ = ∂ ∂ x i + ∂ ∂ y j + ∂ ∂ z k = e i ∂ i \nabla = \frac{\partial }{\partial x} \boldsymbol{i} + \frac{\partial }{\partial y} \boldsymbol{j}+\frac{\partial }{\partial z} \boldsymbol{k} = \bm{e}_i \partial_i
∇ = ∂ x ∂ i + ∂ y ∂ j + ∂ z ∂ k = e i ∂ i
梯度
∇ φ = e i ∂ i φ \nabla\varphi= \bm{e}_i \partial_i \varphi
∇ φ = e i ∂ i φ
散度
∇ ⋅ A = ∂ i A j δ j i = ∂ i A i \nabla \cdot \bm{A} = \partial_i A_j \delta^i_j = \partial_i A_i
∇ ⋅ A = ∂ i A j δ j i = ∂ i A i
Proof
∇ ⋅ A = ( e i ∂ i ) ( ⋅ A j e j ) = ∂ i A j ( e i ⋅ e j ) = ∂ i A j δ j i = ∂ i A i \nabla \cdot \bm{A} = (\bm{e}_i\partial_i) (\cdot A_j \bm{e}_j) = \partial_i A_j (\bm{e}_i \cdot \bm{e}_j) = \partial_i A_j \delta^i_j = \partial_i A_i
∇ ⋅ A = ( e i ∂ i ) ( ⋅ A j e j ) = ∂ i A j ( e i ⋅ e j ) = ∂ i A j δ j i = ∂ i A i
旋度
∇ × A = ε i j k ∂ i A j e k \nabla \times \bm{A} = \varepsilon_{ijk} \partial_i A_j \bm{e}_k
∇ × A = ε i j k ∂ i A j e k
推导过程中，ε i j k , δ j i , e i \varepsilon_{ijk},\delta^i_j,\bm{e}_i ε i j k , δ j i , e i ∂ \partial ∂
偏微分算子满足 ∂ j δ j i = ∂ i \partial_j \delta^i_j = \partial_i ∂ j δ j i = ∂ i
∇ ( A ⋅ B ) = B × ( ∇ × A ) + ( B ⋅ ∇ ) A + A × ( ∇ × B ) + ( A ⋅ ∇ ) B \nabla(\boldsymbol{A} \cdot \boldsymbol{B}) = \bm{B} \times (\nabla \times \bm{A}) + (\bm{B} \cdot \nabla) \bm{A} + \bm{A} \times (\nabla \times \bm{B}) + (\bm{A} \cdot \nabla) \bm{B} ∇ ( A ⋅ B ) = B × ( ∇ × A ) + ( B ⋅ ∇ ) A + A × ( ∇ × B ) + ( A ⋅ ∇ ) B
Proof
B × ( ∇ × A ) = ε i j k B i ( ∇ × A ) j e k = ε i j k B i ε p q j ∂ p A q e k = − ε i k j ε p q j B i ∂ p A q e k = ( δ q i δ p k − δ p i δ q k ) B i ∂ p A q e k = δ q i B i ∂ p A q e p − δ p i B i ∂ p A q e q = δ q i B i ∂ p A q e p − B i ( δ p i ∂ p ) A q e q = δ q i B i ∂ p A q e p − B i ∂ i A q e q = δ q i B i ∂ p A q e p − ( B ⋅ ∇ ) A \begin{aligned}
& \quad\; \boldsymbol{B} \times (\nabla \times \boldsymbol{A}) \\
&= \varepsilon_{ijk}B_i(\nabla \times \boldsymbol{A})_j \boldsymbol{e}_k\\
&= \varepsilon_{ijk}B_i\varepsilon_{pqj} \partial_p A_q \boldsymbol{e}_k \\
&= -\varepsilon_{ikj}\varepsilon_{pqj} B_i \partial_p A_q \boldsymbol{e}_k \\
&= (\delta^i_q\delta^k_p-\delta^i_p \delta^k_q) B_i \partial_p A_q \boldsymbol{e}_k \\
&= \delta^i_q B_i \partial_p A_q \boldsymbol{e}_p - \delta^i_p B_i \partial_p A_q \boldsymbol{e}_q \\
&= \delta^i_q B_i \partial_p A_q \boldsymbol{e}_p - B_i (\delta^i_p \partial_p) A_q \boldsymbol{e}_q \\
&= \delta^i_q B_i \partial_p A_q \boldsymbol{e}_p - B_i \partial_i A_q \boldsymbol{e}_q \\
&= \delta^i_q B_i \partial_p A_q \boldsymbol{e}_p - (\boldsymbol{B} \cdot \nabla) \boldsymbol{A} \\
\end{aligned}
B × ( ∇ × A ) = ε i j k B i ( ∇ × A ) j e k = ε i j k B i ε p q j ∂ p A q e k = − ε i k j ε p q j B i ∂ p A q e k = ( δ q i δ p k − δ p i δ q k ) B i ∂ p A q e k = δ q i B i ∂ p A q e p − δ p i B i ∂ p A q e q = δ q i B i ∂ p A q e p − B i ( δ p i ∂ p ) A q e q = δ q i B i ∂ p A q e p − B i ∂ i A q e q = δ q i B i ∂ p A q e p − ( B ⋅ ∇ ) A
同理有
A × ( ∇ × B ) = δ q i A i ∂ p B q e p − ( A ⋅ ∇ ) B \boldsymbol{A} \times (\nabla \times \boldsymbol{B}) = \delta^i_q A_i \partial_p B_q \boldsymbol{e}_p - (\boldsymbol{A} \cdot \nabla) \boldsymbol{B}
A × ( ∇ × B ) = δ q i A i ∂ p B q e p − ( A ⋅ ∇ ) B
对换 i i i q q q
A × ( ∇ × B ) = δ i q A q ∂ p B i e p − ( A ⋅ ∇ ) B \boldsymbol{A} \times (\nabla \times \boldsymbol{B}) = \delta^q_i A_q \partial_p B_i \boldsymbol{e}_p - (\boldsymbol{A} \cdot \nabla) \boldsymbol{B}
A × ( ∇ × B ) = δ i q A q ∂ p B i e p − ( A ⋅ ∇ ) B
两式相加，并注意到 ∂ p ( B i A q ) = B i ∂ p A q + A q ∂ p B i \partial_p (B_i A_q) = B_i \partial_p A_q + A_q \partial_p B_i ∂ p ( B i A q ) = B i ∂ p A q + A q ∂ p B i
B × ( ∇ × A ) + A × ( ∇ × B ) = δ q i ∂ p ( B i A q ) e p − ( B ⋅ ∇ ) A − ( A ⋅ ∇ ) B = ∂ p ( B i A q δ q i ) e p − ( B ⋅ ∇ ) A − ( A ⋅ ∇ ) B = ∂ p ( B i A i ) e p − ( B ⋅ ∇ ) A − ( A ⋅ ∇ ) B = ∇ ( A ⋅ B ) − ( B ⋅ ∇ ) A − ( A ⋅ ∇ ) B \begin{aligned}
\boldsymbol{B} \times (\nabla \times \boldsymbol{A}) + \boldsymbol{A} \times (\nabla \times \boldsymbol{B}) &= \delta^i_q \partial_p (B_i A_q) \bm{e}_p - (\boldsymbol{B} \cdot \nabla) \boldsymbol{A} - (\boldsymbol{A} \cdot \nabla) \boldsymbol{B} \\
&= \partial_p (B_i A_q \delta^i_q) \bm{e}_p - (\boldsymbol{B} \cdot \nabla) \boldsymbol{A} - (\boldsymbol{A} \cdot \nabla) \boldsymbol{B} \\
&= \partial_p (B_i A_i) \bm{e}_p - (\boldsymbol{B} \cdot \nabla) \boldsymbol{A} - (\boldsymbol{A} \cdot \nabla) \boldsymbol{B} \\
&= \nabla(\boldsymbol{A} \cdot \boldsymbol{B}) - (\boldsymbol{B} \cdot \nabla) \boldsymbol{A} - (\boldsymbol{A} \cdot \nabla) \boldsymbol{B} \\
\end{aligned}
B × ( ∇ × A ) + A × ( ∇ × B ) = δ q i ∂ p ( B i A q ) e p − ( B ⋅ ∇ ) A − ( A ⋅ ∇ ) B = ∂ p ( B i A q δ q i ) e p − ( B ⋅ ∇ ) A − ( A ⋅ ∇ ) B = ∂ p ( B i A i ) e p − ( B ⋅ ∇ ) A − ( A ⋅ ∇ ) B = ∇ ( A ⋅ B ) − ( B ⋅ ∇ ) A − ( A ⋅ ∇ ) B
证毕
∇ × ( A × B ) = ( B ⋅ ∇ ) A − ( ∇ ⋅ A ) B + ( ∇ ⋅ B ) A − ( A ⋅ ∇ ) B \nabla \times (\bm{A} \times \bm{B}) = (\bm{B} \cdot \nabla) \bm{A} - (\nabla \cdot \bm{A}) \bm{B} + (\nabla \cdot \bm{B}) \bm{A} - (\bm{A} \cdot \nabla) \bm{B} ∇ × ( A × B ) = ( B ⋅ ∇ ) A − ( ∇ ⋅ A ) B + ( ∇ ⋅ B ) A − ( A ⋅ ∇ ) B
Proof
∇ × ( A × B ) = ε i j k ∂ i ( A × B ) j e k = ε i j k ∂ i ε p q j A p B q e k = ( δ q i δ p k − δ p i δ q k ) ∂ i ( A p B q ) e k = ( δ q i δ p k − δ p i δ q k ) ( A p ∂ i B q + B q ∂ i A p ) e k \begin{aligned}
& \quad\; \nabla \times (\bm{A} \times \bm{B}) \\
&= \varepsilon_{ijk} \partial_i (\bm{A} \times \bm{B})_j \bm{e}_k \\
&= \varepsilon_{ijk} \partial_i \varepsilon_{pqj} A_p B_q \bm{e}_k \\
&= (\delta^i_q\delta^k_p-\delta^i_p \delta^k_q) \partial_i (A_p B_q) \bm{e}_k \\
&= (\delta^i_q\delta^k_p-\delta^i_p \delta^k_q)(A_p \partial_i B_q + B_q \partial_i A_p) \bm{e}_k
\end{aligned}
∇ × ( A × B ) = ε i j k ∂ i ( A × B ) j e k = ε i j k ∂ i ε p q j A p B q e k = ( δ q i δ p k − δ p i δ q k ) ∂ i ( A p B q ) e k = ( δ q i δ p k − δ p i δ q k ) ( A p ∂ i B q + B q ∂ i A p ) e k
展开得到四项，它们分别是
δ q i δ p k A p ∂ i B q e k = A p e p ∂ p B q = A ( ∇ ⋅ B ) \delta^i_q\delta^k_p A_p \partial_i B_q \boldsymbol{e}_k = A_p \boldsymbol{e}_p \partial_p B_q = \bm{A} (\nabla \cdot \bm{B})
δ q i δ p k A p ∂ i B q e k = A p e p ∂ p B q = A ( ∇ ⋅ B )
− δ p i δ q k A p ∂ i B q e k = − A p ∂ p B q e q = − ( A ⋅ ∇ ) B -\delta^i_p \delta^k_q A_p \partial_i B_q \bm{e}_k = -A_p \partial_p B_q \bm{e}_q = -(\boldsymbol{A} \cdot \nabla) \boldsymbol{B}
− δ p i δ q k A p ∂ i B q e k = − A p ∂ p B q e q = − ( A ⋅ ∇ ) B
δ q i δ p k B q ∂ i A p e k = B q ∂ q A p e p = ( B ⋅ ∇ ) A \delta^i_q\delta^k_p B_q \partial_i A_p \bm{e_k} = B_q \partial_q A_p \bm{e}_p = (\bm{B} \cdot \nabla) \bm{A}
δ q i δ p k B q ∂ i A p e k = B q ∂ q A p e p = ( B ⋅ ∇ ) A
− δ p i δ q k B q ∂ i A p e k = − B q e q ∂ p A p = − ( ∇ ⋅ A ) B -\delta^i_p \delta^k_q B_q \partial_i A_p \bm{e}_k = -B_q \bm{e}_q \partial_p A_p = - (\nabla \cdot \bm{A}) \bm{B}
− δ p i δ q k B q ∂ i A p e k = − B q e q ∂ p A p = − ( ∇ ⋅ A ) B
证毕
