矢量分析 关于∇一些矢量恒等式速记法-CSDN博客 (2024)

前言

20200413
最近在学电磁场与电磁波,那几个有 ∇ \nabla 的矢量恒等式真的够呛啊…
在B站上看到的矢量分析视频, 该大神UP的理解简直到位…

∇ \nabla 看成是有矢量性微分性的东东

公式基础

先回忆几个公式
A × ( B × C ) = B ( A ⋅ C ) − ( A ⋅ B ) C \mathbf{A\times (B\times C)=B(A\cdot C)-(A\cdot B)C} A×(B×C)=B(AC)(AB)C
为了好记可以写成
A × ( B × C ) + ( A ⋅ B ) C = B ( A ⋅ C ) \mathbf{A\times (B\times C)+(A\cdot B)C=B(A\cdot C)} A×(B×C)+(AB)C=B(AC)
记忆方法:左边都是ABC,左边的括号中,一个是BC,一个是AB,那剩下的一个括号肯定就是AC咯

混合积:
A ⋅ ( B × C ) = C ⋅ ( A × B ) = B ⋅ ( C × A ) \mathbf{A\cdot (B\times C)}=\mathbf{C\cdot (A\times B)}=\mathbf{B\cdot (C\times A)} A(B×C)=C(A×B)=B(C×A)

列举出有关 ∇ \nabla 的恒等式

排列组合看看有哪些公式:

首先需要明确维度的问题

∇ \nabla 是梯度, 一般作用于标量升维为矢量
∇ ⋅ \nabla\cdot 是散度, 一般作用于矢量降维为标量
∇ × \nabla\times ×是旋度, 一般作用于矢量既不升维为矩阵也不降维为标量,保持原来的矢量状态

对于2个 ∇ \nabla 作用于一个矢量函数的(不考虑矢量升维、标量降维)

最前面有(变成矢量或标量)一般没有没有的原因一般也没有没有的原因
∇ ⋅ \nabla\cdot ∇ ⋅ ( ∇ × A ) \nabla\cdot(\nabla\times\mathbf{A}) (×A) ∇ ⋅ ( ∇ ⋅ A ) \nabla\cdot(\nabla\cdot\mathbf{A}) (A)散度给标量降维 ∇ ⋅ ( ∇ A ) \nabla\cdot(\nabla\mathbf{A}) (A)矢量升上的张量不会算
∇ × \nabla\times × ∇ × ∇ ψ \nabla\times\nabla\psi ×ψ ∇ × ( ∇ ⋅ ψ ) \nabla\times(\nabla\cdot\psi) ×(ψ)散度给标量降维 ∇ × ( ∇ × ψ ) \nabla\times(\nabla\times\psi) ×(×ψ)标量求旋度是啥玩意?
∇ × \nabla\times × ∇ × ∇ × A \nabla\times\nabla\times \mathbf{A} ××A ∇ × ( ∇ ⋅ A ) \nabla\times(\nabla\cdot \mathbf{A}) ×(A)给标量求旋度了 ∇ × ( ∇ A ) \nabla\times(\nabla \mathbf{A}) ×(A)最终得到升维的张量
∇ \nabla ∇ ∇ ⋅ A = ∇ 2 A \nabla\nabla\cdot\mathbf{A}=\nabla^2\mathbf{A} A=2A ∇ ( ∇ A ) \nabla(\nabla\mathbf{A}) (A)矢量升了2个维度 ∇ ( ∇ × A ) \nabla(\nabla\times\mathbf{A}) (×A)最终得到升维的张量

对于1个 ∇ \nabla 作用于两个矢量函数的

∇ ⋅ ( A × B ) \nabla\cdot(\mathbf{A\times B}) (A×B)
∇ × ( A × B ) \nabla\times(\mathbf{A\times B}) ×(A×B)
∇ ( A ⋅ B ) \nabla(\mathbf{A\cdot B}) (AB)
∇ ( ψ ϕ ) \nabla(\psi\phi) (ψϕ)

推导开始

只体现矢量性

在接下来的式子中, ∇ \nabla 只体现矢量性,因为式子中只存在一个矢量函数,除了它没别的能微分了,所以只有矢量性

1. ∇ ⋅ ( ∇ × A ) = 0 \nabla\cdot(\nabla\times\mathbf{A})=0 (×A)=0

由于 ∇ \nabla 和自己垂直的 ∇ \nabla (即 ∇ × A \nabla\times\mathbf{A} ×A)点乘, 结果一定为0

2. ∇ × ∇ ψ = 0 \nabla\times\nabla\psi=0 ×ψ=0

由于 ∇ \nabla 和自己叉乘,结果一定为0

3. ∇ × ∇ × A = ∇ ∇ ⋅ A − ∇ ⋅ ∇ A \nabla\times\nabla\times \mathbf{A}=\nabla\nabla\cdot\mathbf{A}-\nabla\cdot\nabla\mathbf{A} ××A=AA

因为 A × ( B × C ) = B ( A ⋅ C ) − ( A ⋅ B ) C \mathbf{A\times (B\times C)=B(A\cdot C)-(A\cdot B)C} A×(B×C)=B(AC)(AB)C
(注意这里把C都放在最后,并且向量和标量相乘时不要用点号,直接"隐形乘法")
对比一下公式,会得到
∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇ ⋅ ( ∇ A ) \nabla\times(\nabla\times \mathbf{A})=\nabla(\nabla\cdot\mathbf{A})-\nabla\cdot(\nabla\mathbf{A}) ×(×A)=(A)(A)
因为 ∇ \nabla 看成矢量, 矢量的点乘, 数乘所以括号可以随意拆随意加(实际上得到的公式如果不按加括号的来算,大家可能都不会算hhh)
然后 ∇ \nabla 的叉乘 ∇ × ∇ × A \nabla\times\nabla\times \mathbf{A} ××A,可以在后面加括号的原因是,:
如果先算前面,这时你会发现到底应该是 ( ∇ × ∇ ) × A (\nabla\times\nabla)\times \mathbf{A} (×)×A还是 ∇ × ( ∇ × A ) \nabla\times(\nabla\times\mathbf{A}) ×(×A), 撇开这个,最重要的是 ∇ × ∇ \nabla\times\nabla ×肯定为0啊,因为自己叉乘自己,所以式子没有意义, 那就默认是在后面加括号了,更加肯定我们的想法.

∇ × ∇ × A = ∇ × ( ∇ × A ) = ∇ ∇ ⋅ A − ∇ 2 A = ∇ ∇ ⋅ A − ∇ ⋅ ∇ A = ∇ ( ∇ ⋅ A ) − ∇ ⋅ ( ∇ A ) \nabla\times\nabla\times \mathbf{A}=\nabla\times(\nabla\times\mathbf{A})=\nabla\nabla\cdot\mathbf{A}-\nabla^2\mathbf{A}=\nabla\nabla\cdot\mathbf{A}-\nabla\cdot\nabla\mathbf{A}=\nabla(\nabla\cdot\mathbf{A})-\nabla\cdot(\nabla\mathbf{A}) ××A=×(×A)=A2A=AA=(A)(A)

体现矢量性和微分性----中心思想 ∇ = ∇ A + ∇ B \nabla=\nabla_A+\nabla_B =A+B

下面的式子关于多个矢量函数,所以微分性就体现出来了,具体体现在对 A \mathbf{A} A做微分的同时,也要对 B \mathbf{B} B做微分, 也即 ∇ = ∇ A + ∇ B \nabla=\nabla_A+\nabla_B =A+B

1. ∇ ( ψ ϕ ) = ψ ∇ ϕ + ψ ∇ ϕ \nabla(\psi\phi)=\psi\nabla\phi+\psi\nabla\phi (ψϕ)=ψϕ+ψϕ

这个很简单呀
∇ ( ψ ϕ ) = ( ∇ ψ + ∇ ϕ ) ( ψ ϕ ) \nabla(\psi\phi)=(\nabla_\psi+\nabla_\phi)(\psi\phi) (ψϕ)=(ψ+ϕ)(ψϕ)
= ∇ ψ ( ϕ ψ ) + ∇ ϕ ( ψ ϕ ) =\nabla_\psi(\phi\psi)+\nabla_\phi(\psi\phi) =ψ(ϕψ)+ϕ(ψϕ)
= ϕ ∇ ψ ψ + ψ ∇ ϕ ϕ =\phi\nabla_\psi\psi+\psi\nabla_\phi\phi =ϕψψ+ψϕϕ
= ϕ ∇ ψ + ψ ∇ ϕ =\phi\nabla\psi+\psi\nabla\phi =ϕψ+ψϕ

完成

再看个不复杂的

2. ∇ ⋅ ( A × B ) = B ⋅ ( ∇ × A ) − A ⋅ ( ∇ × B ) \nabla\cdot(\mathbf{A\times B})=\mathbf{B}\cdot(\nabla\times \mathbf{A)-A}\cdot(\nabla\times \mathbf{B}) (A×B)=B(×A)A(×B)

注意到这不就是混合积嘛, A ⋅ ( B × C ) = C ⋅ ( A × B ) = B ⋅ ( C × A ) \mathbf{A\cdot (B\times C)}=\mathbf{C\cdot (A\times B)}=\mathbf{B\cdot (C\times A)} A(B×C)=C(A×B)=B(C×A)

但是按理说只会有一项 B ⋅ ∇ × A \mathbf{B}\cdot\nabla\times \mathbf{A} B×A呀,怎么会多出一项 − A ⋅ ∇ × B -\mathbf{A}\cdot\nabla\times \mathbf{B} A×B呢?
原因就是在 ∇ \nabla 的微分性, 对两个矢量函数都要微分
B ⋅ ∇ × A \mathbf{B}\cdot\nabla\times \mathbf{A} B×A A \mathbf{A} A微分了,还应该加上对 B \mathbf{B} B微分的一项
因此,根据轮换, 应该得到 A ⋅ ( B × ∇ ) \mathbf{A}\cdot(\mathbf{B}\times\nabla) A(B×)
到这里, 你傻眼了, B × ∇ \mathbf{B}\times\nabla B×是什么
实际上又回到矢量性了, 因为从 B × ∇ \mathbf{B}\times\nabla B×的角度看, 只有一个矢量函数 B \mathbf{B} B
因此, 看成矢量的话, B × ∇ = − ∇ × B \mathbf{B}\times\nabla=-\nabla\times\mathbf{B} B×=×B
所以, 搞定
∇ = ∇ A + ∇ B \nabla=\nabla_A+\nabla_B =A+B看待的话,就是如下推导

∇ ⋅ ( A × B ) = ( ∇ A + ∇ B ) ⋅ ( A × B ) \nabla\cdot(\mathbf{A\times B})=(\nabla_A+\nabla_B)\cdot(\mathbf{A\times B}) (A×B)=(A+B)(A×B)
= ∇ A ⋅ ( A × B ) + ∇ B ⋅ ( A × B ) =\nabla_A\cdot(\mathbf{A\times B})+\nabla_B\cdot(\mathbf{A\times B}) =A(A×B)+B(A×B)
= B ⋅ ∇ A × A + A ⋅ ( B × ∇ B ) =\mathbf{B}\cdot\nabla_A\times \mathbf{A}+\mathbf{A}\cdot(\mathbf{B}\times\nabla_B) =BA×A+A(B×B)
= B ⋅ ∇ A × A − A ⋅ ( ∇ B × B ) =\mathbf{B}\cdot\nabla_A\times \mathbf{A}-\mathbf{A}\cdot(\nabla_B\times\mathbf{B}) =BA×AA(B×B)
= B ⋅ ∇ × A − A ⋅ ( ∇ × B ) =\mathbf{B}\cdot\nabla\times \mathbf{A}-\mathbf{A}\cdot(\nabla\times\mathbf{B}) =B×AA(×B)

\

3. ∇ × ( A × B ) = A ( ∇ ⋅ B ) − B ( ∇ ⋅ A ) + ( B ⋅ ∇ ) A − ( A ⋅ ∇ ) B \nabla\times(\mathbf{A\times B})=\mathbf{A}(\nabla\cdot\mathbf{B)}-\mathbf{B}(\nabla\cdot\mathbf{A})+(\mathbf{B}\cdot\nabla)\mathbf{A}-(\mathbf{A}\cdot\nabla)\mathbf{B} ×(A×B)=A(B)B(A)+(B)A(A)B

由于
A × ( B × C ) = B ( A ⋅ C ) − C ( A ⋅ B ) \mathbf{A\times (B\times C)=B(A\cdot C)-C(A\cdot B)} A×(B×C)=B(AC)C(AB)
同理
∇ × ( A × B ) = ( ∇ A + ∇ B ) × ( A × B ) \nabla\times(\mathbf{A\times B})=(\nabla_A+\nabla_B)\times(\mathbf{A\times B}) ×(A×B)=(A+B)×(A×B)
= ∇ A × ( A × B ) + ∇ B × ( A × B ) =\nabla_A\times(\mathbf{A\times B})+\nabla_B\times(\mathbf{A\times B}) =A×(A×B)+B×(A×B)
= A ( ∇ A ⋅ B ) − B ( ∇ A ⋅ A ) + A ( ∇ B ⋅ B ) − B ( ∇ B ⋅ A ) =\mathbf{A}(\nabla_A\cdot\mathbf{B})-\mathbf{B}(\nabla_A\cdot\mathbf{A})+\mathbf{A}(\nabla_B\cdot\mathbf{B})-\mathbf{B}(\nabla_B\cdot\mathbf{A}) =A(AB)B(AA)+A(BB)B(BA)
注意到项 A ( ∇ A ⋅ B ) \mathbf{A}(\nabla_A\cdot\mathbf{B}) A(AB) − B ( ∇ B ⋅ A ) -\mathbf{B}(\nabla_B\cdot\mathbf{A}) B(BA), 实际上只是公式的位置反了, ∇ A \nabla_A A应该是在 A \mathbf{A} A的前面, 并且把 ∇ A \nabla_A A看成矢量的话, 点乘可以交换顺序, 因此
A ( ∇ A ⋅ B ) = ( B ⋅ ∇ A ) A \mathbf{A}(\nabla_A\cdot\mathbf{B})=(\mathbf{B}\cdot\nabla_A)\mathbf{A} A(AB)=(BA)A,
B ( ∇ B ⋅ A ) = ( A ⋅ ∇ B ) B \mathbf{B}(\nabla_B\cdot\mathbf{A})=(\mathbf{A}\cdot\nabla_B)\mathbf{B} B(BA)=(AB)B
= ( B ⋅ ∇ A ) A − B ( ∇ A ⋅ A ) + A ( ∇ B ⋅ B ) − ( A ⋅ ∇ B ) B =(\mathbf{B}\cdot\nabla_A)\mathbf{A}-\mathbf{B}(\nabla_A\cdot\mathbf{A})+\mathbf{A}(\nabla_B\cdot\mathbf{B})-(\mathbf{A}\cdot\nabla_B)\mathbf{B} =(BA)AB(AA)+A(BB)(AB)B
= A ( ∇ B ⋅ B ) − B ( ∇ A ⋅ A ) + ( A ⋅ ∇ B ) B − B ( ∇ B ⋅ A ) =\mathbf{A}(\nabla_B\cdot\mathbf{B})-\mathbf{B}(\nabla_A\cdot\mathbf{A})+(\mathbf{A}\cdot\nabla_B)\mathbf{B}-\mathbf{B}(\nabla_B\cdot\mathbf{A}) =A(BB)B(AA)+(AB)BB(BA)
完成

\

4. ∇ ( A ⋅ B ) = ( B ⋅ ∇ ) A + ( A ⋅ ∇ ) B + B × ( ∇ × A ) + A × ( ∇ × B ) \nabla(\mathbf{A\cdot B})=(\mathbf{B}\cdot\nabla)\mathbf{A}+(\mathbf{A}\cdot\nabla)\mathbf{B}+\mathbf{B}\times(\nabla\times\mathbf{A})+\mathbf{A}\times(\nabla\times\mathbf{B}) (AB)=(B)A+(A)B+B×(×A)+A×(×B)

还是用这个公式 A × ( B × C ) = B ( A ⋅ C ) − C ( A ⋅ B ) \mathbf{A\times (B\times C)=B(A\cdot C)-C(A\cdot B)} A×(B×C)=B(AC)C(AB)
只不过交换一下位置
B ( A ⋅ C ) = C ( A ⋅ B ) + A × ( B × C ) \mathbf{B(A\cdot C)=C(A\cdot B)+A\times (B\times C)} B(AC)=C(AB)+A×(B×C)


∇ ( A ⋅ B ) = ( ∇ A + ∇ B ) ( A ⋅ B ) \nabla(\mathbf{A\cdot B})=(\nabla_A+\nabla_B)(\mathbf{A\cdot B}) (AB)=(A+B)(AB)

我们要让被微分的放到最后, 才能使得一些运算合理, 因为对A微分,相当于B是常数, 要放前面较合理
= ∇ A ( B ⋅ A ) + ∇ B ( A ⋅ B ) =\nabla_A(\mathbf{B\cdot A})+\nabla_B(\mathbf{A\cdot B}) =A(BA)+B(AB)
= A ( B ⋅ ∇ A ) + B × ( ∇ A × A ) + B ( A ⋅ ∇ B ) + A × ( ∇ B × B ) =\mathbf{A}(\mathbf{B}\cdot\nabla_A)+\mathbf{B}\times(\nabla_A\times\mathbf{A})+\mathbf{B}(\mathbf{A}\cdot\nabla_B)+\mathbf{A}\times(\nabla_B\times\mathbf{B}) =A(BA)+B×(A×A)+B(AB)+A×(B×B)
交换 ∇ \nabla 的位置
= ( B ⋅ ∇ A ) A + B × ( ∇ A × A ) + ( A ⋅ ∇ B ) B + A × ( ∇ B × B ) =(\mathbf{B}\cdot\nabla_A)\mathbf{A}+\mathbf{B}\times(\nabla_A\times\mathbf{A})+(\mathbf{A}\cdot\nabla_B)\mathbf{B}+\mathbf{A}\times(\nabla_B\times\mathbf{B}) =(BA)A+B×(A×A)+(AB)B+A×(B×B)
交换2 3项之间的位置
= ( B ⋅ ∇ A ) A + ( A ⋅ ∇ B ) B + B × ( ∇ A × A ) + A × ( ∇ B × B ) =(\mathbf{B}\cdot\nabla_A)\mathbf{A}+(\mathbf{A}\cdot\nabla_B)\mathbf{B}+\mathbf{B}\times(\nabla_A\times\mathbf{A})+\mathbf{A}\times(\nabla_B\times\mathbf{B}) =(BA)A+(AB)B+B×(A×A)+A×(B×B)
完成

总结

这些只是一个用于记忆的小技巧, 不要太深究里面的数学严谨性, 算是一种比较深刻的理解把, 就像一些直觉虽然不严谨, 但是它的的确确能得到结果. 希望能帮到大家.
这些推导看起来实际上真的很无聊, 但是自己动笔推一边过后, 就会记得非常非常清楚

注意:上述的骚操作只对提及的公式有效,其他公式不能保证

后记

PS:这是作者的第一篇blog, 排版属实丑,见谅见谅
这是我在学Einstein求和约定时学到的, 有机会再总结一下那个UP的Einstein求和约定吧, 超强的东西

最后附上参考的链接

参考文献

https://www.bilibili.com/video/BV1VW41127Dd?p=7

矢量分析 关于∇一些矢量恒等式速记法-CSDN博客 (2024)
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