中南林业科技大学理学院，湖南 长沙

收稿日期：2021年5月13日；录用日期：2021年6月18日；发布日期：2021年6月28日

摘要

电动力学教材一般仅仅在附录部分给出曲线正交坐标系下标量的梯度 $\stackrel{\to }{\nabla }\psi$ 、矢量的散度 $\stackrel{\to }{\nabla }\cdot \stackrel{\to }{A}$ 和旋度 $\stackrel{\to }{\nabla }\times \stackrel{\to }{A}$，以及标量的拉普拉斯算符 ${\nabla }^2\psi$ 的一般表达式。在教学中我们发现， $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ 和 ${\nabla }^2\stackrel{\to }{A}$ 在曲线正交坐标系下的表达式也是很重要的。本文从微分几何的角度推导出三维欧氏空间中曲线正交坐标系下 $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ 和 ${\nabla }^2\stackrel{\to }{A}$ 的一般表达式，以及它们在直角坐标系、柱坐标系和球坐标系的具体表达式，供感兴趣的学生和教师参考。

关键词

曲线正交坐标系，微分几何， $\stackrel{\to }{\nabla }\stackrel{\to }{A}$，${\nabla }^2\stackrel{\to }{A}$

A Derivation of $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ and ${\nabla }^2\stackrel{\to }{A}$ in Curvilinear Orthogonal Coordinates

Gerui Chen^*, Mengdong He, Xiaoyan Chen

College of Science, Central South University of Forestry and Technology, Changsha Hunan

Received: May 13^th, 2021; accepted: Jun. 18^th, 2021; published: Jun. 28^th, 2021

ABSTRACT

In textbooks of electrodynamics, the formulas of the gradient $\stackrel{\to }{\nabla }\psi$, divergence $\stackrel{\to }{\nabla }\cdot \stackrel{\to }{A}$, curl $\stackrel{\to }{\nabla }\times \stackrel{\to }{A}$，and Laplace operator ${\nabla }^2\psi$ in curvilinear orthogonal coordinates are often given in the appendix. However, in the teaching process, we find that the expressions of $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ and ${\nabla }^2\stackrel{\to }{A}$ in curvilinear orthogonal coordinates are also very important. In this paper, the general expressions of $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ and ${\nabla }^2\stackrel{\to }{A}$ in curvilinear orthogonal coordinates in three-dimensional Euclidean space are derived based on the method of differential geometry. And the formulas of $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ and ${\nabla }^2\stackrel{\to }{A}$ in the Cartesian coordinates, cylindrical coordinates and spherical coordinates are also given for convenience of interested students and teachers.

Keywords:Curvilinear Orthogonal Coordinates, Differential Geometry, $\stackrel{\to }{\nabla }\stackrel{\to }{A}$,${\nabla }^2\stackrel{\to }{A}$

Copyright © 2021 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

在电动力学课程中，一般在讲授正式内容以前要先讲授教材附录部分中如下8个 $\stackrel{\to }{\nabla }$ 算符运算公式的证明，因为它们在后面的课程中要频繁用到 [1]。

$\stackrel{\to }{\nabla }\left(\phi \psi \right)=\phi \stackrel{\to }{\nabla }\psi +\psi \stackrel{\to }{\nabla }\phi$ (1)

$\stackrel{\to }{\nabla }\cdot \left(\phi \stackrel{\to }{A}\right)=\left(\stackrel{\to }{\nabla }\phi \right)\cdot \stackrel{\to }{A}+\phi \stackrel{\to }{\nabla }\cdot \stackrel{\to }{A}$ (2)

$\stackrel{\to }{\nabla }\times \left(\phi \stackrel{\to }{A}\right)=\left(\stackrel{\to }{\nabla }\phi \right)\times \stackrel{\to }{A}+\phi \stackrel{\to }{\nabla }\times \stackrel{\to }{A}$ (3)

$\stackrel{\to }{\nabla }\cdot \left(\stackrel{\to }{A}\times \stackrel{\to }{B}\right)=\left(\stackrel{\to }{\nabla }\times \stackrel{\to }{A}\right)\cdot \stackrel{\to }{B}-\stackrel{\to }{A}\cdot \left(\stackrel{\to }{\nabla }\times \stackrel{\to }{B}\right)$ (4)

$\stackrel{\to }{\nabla }\times \left(\stackrel{\to }{A}\times \stackrel{\to }{B}\right)=\left(\stackrel{\to }{B}\cdot \stackrel{\to }{\nabla }\right)\stackrel{\to }{A}+\left(\stackrel{\to }{\nabla }\cdot \stackrel{\to }{B}\right)\stackrel{\to }{A}-\left(\stackrel{\to }{A}\cdot \stackrel{\to }{\nabla }\right)\stackrel{\to }{B}-\left(\stackrel{\to }{\nabla }\cdot \stackrel{\to }{A}\right)\stackrel{\to }{B}$ (5)

$\stackrel{\to }{\nabla }\left(\stackrel{\to }{A}\cdot \stackrel{\to }{B}\right)=\stackrel{\to }{A}\times \left(\stackrel{\to }{\nabla }\times \stackrel{\to }{B}\right)+\left(\stackrel{\to }{A}\cdot \stackrel{\to }{\nabla }\right)\stackrel{\to }{B}+\stackrel{\to }{B}\times \left(\stackrel{\to }{\nabla }\times \stackrel{\to }{A}\right)+\left(\stackrel{\to }{B}\cdot \stackrel{\to }{\nabla }\right)\stackrel{\to }{A}$ (6)

$\stackrel{\to }{\nabla }\cdot \stackrel{\to }{\nabla }\phi ={\nabla }^2\phi$ (7)

$\stackrel{\to }{\nabla }\times \left(\stackrel{\to }{\nabla }\times \stackrel{\to }{A}\right)=\stackrel{\to }{\nabla }\left(\stackrel{\to }{\nabla }\cdot \stackrel{\to }{A}\right)-{\nabla }^2\stackrel{\to }{A}$ (8)

电动力学课程中，除了直角坐标系，还经常用到柱坐标系和球坐标系等曲线正交坐标系。教材一般在附录部分不加证明地给出曲线正交坐标系下标量的梯度 $\stackrel{\to }{\nabla }\psi$ 、矢量的散度 $\stackrel{\to }{\nabla }\cdot \stackrel{\to }{A}$ 和旋度 $\stackrel{\to }{\nabla }\times \stackrel{\to }{A}$，以及标量的拉普拉斯算符 ${\nabla }^2\psi$ 的一般表达式。然而仅仅利用这四个式子，上面的第(5)、(6)、(8)式却无法求出，因为里面涉及到 $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ 与 ${\nabla }^2\stackrel{\to }{A}$ 的运算。上面第(5)、(6)式需要利用 $\stackrel{\to }{\nabla }\stackrel{\to }{A}$，而第(8)式需要利用 ${\nabla }^2\stackrel{\to }{A}$。电动力学课程中，电磁波的传播是重要的一章。时谐电磁波满足亥姆赫兹方程 ${\nabla }^2\stackrel{\to }{E}+k\stackrel{\to }{E}=0$，解此偏微分方程需要根据问题的对称性选择合适的坐标系。教材中讨论矩形谐振腔和波导，选择直角坐标系方便，但是如果谐振腔和波导具有球对称性和柱对称性，则需要 ${\nabla }^2\stackrel{\to }{E}$ 在球坐标系和柱坐标系的表达式。由此可见 $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ 与 ${\nabla }^2\stackrel{\to }{A}$ 在曲线正交坐标系下的表达式是重要的。本文从微分几何的角度推导三维欧氏空间中曲线正交坐标系下 $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ 与 ${\nabla }^2\stackrel{\to }{A}$ 的一般表达式。微分几何虽然是数学系本科生学习的内容，但是对学过高等数学和线性代数的二年级本科生在教师的指点下也可以大致掌握基本的思想和方法 [2]。作者讲授应用物理专业电动力学课程，课程最后要讲授狭义相对论，学生学完后就具有了一定的几何思想。有不少同学对狭义相对论之后的广义相对论兴趣十分浓厚，往往希望教师能够给出一些学习建议。近代物理的重要内容广义相对论和规范场论都是建立在整体微分几何的基础上的。鉴于近代微分几何对物理工作者的重要性，我们认为在本科教学中适当指导学生了解一点微分几何是极有帮助的，这对激发学生探索物理的兴趣起到很大的作用。文中详细展现每一步的推导计算过程，这样有助于读者初步了解微分几何的思想和方法。本文的计算采用著作 [2] 的符号约定。

2. 准备工作

三维欧氏空间在笛卡尔坐标系下的线元为：

$\text{d}{l}^2=\text{d}x\text{d}x+\text{d}y\text{d}y+\text{d}z\text{d}z$ (9)

设曲线正交坐标系的坐标为 $u_i\left(i=1,2,3\right)$，有变换关系：

$x=x\left(u_1,u_2,u_3\right)$，$y=y\left(u_1,u_2,u_3\right)$，$z=z\left(u_1,u_2,u_3\right)$ (10)

对x、y、z求全微分，带入线元(9)得到三维欧氏空间在曲线正交坐标系下的线元：

$\text{d}{l}^2=h_1^2\text{d}{u}_1\text{d}{u}_1+h_2^2\text{d}{u}_2\text{d}{u}_2+h_3^2\text{d}{u}_3\text{d}{u}_3$ (11)

其中：

$h_i^2={\left(\frac{\partial x}{\partial u_i}\right)}^2+{\left(\frac{\partial y}{\partial u_i}\right)}^2+{\left(\frac{\partial z}{\partial u_i}\right)}^2$，其中 $i=1,2,3$ (12)

为标度因子。

由曲线正交坐标系下的线元表达式(11)，可得度规：

$g_{ij}=\left(\begin{array}{ccc}{h}_1^2& 0& 0\\ 0& h_2^2& 0\\ 0& 0& h_3^2\end{array}\right)$，其中 $i,j=1,2,3$ (13)

及度规的逆：

$g^{ij}=\left(\begin{array}{ccc}\frac{1}{h_1^2}& 0& 0\\ 0& \frac{1}{h_2^2}& 0\\ 0& 0& \frac{1}{h_3^2}\end{array}\right)$，其中 $i,j=1,2,3$ (14)

我们采用罗杰·彭罗斯¹的抽象指标记号，矢量 $\stackrel{\to }{A}$ 用记号 $A^a$ 表示，它在曲线正交坐标系的正交归一基底 ${\left(e_i\right)}^a$ (即 ${\stackrel{\to }{e}}_i$ )下可表示为：

$A^a=A^1{\left(e_1\right)}^a+A^2{\left(e_2\right)}^a+A^3{\left(e_3\right)}^a$ (15)

其中 $A^i\left(i=1,2,3\right)$ 是 $A^a$ 在正交归一基底的分量。

矢量 $A^a$ 在坐标基底 ${\left(\frac{\partial }{\partial u_i}\right)}^a$ 下可表示为：

$A^a=A^{u_1}{\left(\frac{\partial }{\partial u_1}\right)}^a+A^{u_2}{\left(\frac{\partial }{\partial u_2}\right)}^a+A^{u_3}{\left(\frac{\partial }{\partial u_3}\right)}^a$ (16)

其中 $A^{u_i}\left(i=1,2,3\right)$ 是矢量 $A^a$ 在坐标基底的分量。

注意正交归一基底和坐标基底有如下的关系：

${\left(e_i\right)}^a=\frac{1}{h_i}{\left(\frac{\partial }{\partial u_i}\right)}^a$ (17)

矢量 $A^a$ 在正交归一基底的分量 $A^i$ 与坐标基底的分量 $A^{u_i}$ 有关系：

$A^{u_i}=\frac{A^i}{h_i}$ (18)

3. $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ 的计算

$\stackrel{\to }{\nabla }\stackrel{\to }{A}$ 用抽象指标可表示为 $\stackrel{\to }{\nabla }\stackrel{\to }{A}={\nabla }^a{A}^b$，则：

$\stackrel{\to }{\nabla }\stackrel{\to }{A}={\nabla }^a{A}^b=g^{ac}{\nabla }_c{A}^b=g^{ac}\left({\partial }_c{A}^b+{\Gamma }_{ce}^b{A}^e\right)=g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e$ (19)

${\nabla }_a$ 是跟三维欧氏度规相适配的导数算符，与曲线正交坐标系的导数算符 ${\partial }_a$ 有下面的关系：

${\nabla }_c{A}^b={\partial }_c{A}^b+{\Gamma }_{ce}^b{A}^e$ (20)

其中 ${\Gamma }_{ab}^d$ 是反映 ${\nabla }_a$ 与 ${\partial }_a$ 区别的克氏符，注意克氏符下面的两个指标是对称的，即 ${\Gamma }_{ab}^d={\Gamma }_{ba}^d$。

根据克氏符的定义 ${\Gamma }_{\mu \nu }^{\sigma }=\frac{1}{2}{g}^{\sigma \rho }\left(g_{\rho \mu ,\nu }+g_{\nu \rho ,\mu }-g_{\mu \nu ,\rho }\right)$，利用度规(13)以及度规的逆(14)可以算出非零克氏符，如下²

${\Gamma }_{11}^1=\frac{{\partial }_1{h}_1}{h_1}$，${\Gamma }_{12}^1={\Gamma }_{21}^1=\frac{{\partial }_2{h}_1}{h_1}$，${\Gamma }_{13}^1=\frac{{\partial }_3{h}_1}{h_1}$，${\Gamma }_{22}^1=-\frac{h_2{\partial }_1{h}_2}{h_1^2}$，${\Gamma }_{31}^1=\frac{{\partial }_3{h}_1}{h_1}$，${\Gamma }_{33}^1=-\frac{h_3{\partial }_1{h}_3}{h_1^2}$，

${\Gamma }_{11}^2=-\frac{h_1{\partial }_2{h}_1}{h_2^2}$，${\Gamma }_{12}^2={\Gamma }_{21}^2=\frac{{\partial }_1{h}_2}{h_2}$，${\Gamma }_{22}^2=\frac{{\partial }_2{h}_2}{h_2}$，${\Gamma }_{32}^2=\frac{{\partial }_3{h}_2}{h_2}$，${\Gamma }_{33}^2=-\frac{h_3{\partial }_2{h}_3}{h_2^2}$，${\Gamma }_{11}^3=-\frac{h_1{\partial }_3{h}_1}{h_3^2}$ (21)

${\Gamma }_{13}^3={\Gamma }_{31}^3=\frac{{\partial }_1{h}_3}{h_3}$，${\Gamma }_{22}^3=-\frac{h_2{\partial }_3{h}_2}{h_3^2}$，${\Gamma }_{23}^3={\Gamma }_{32}^3=\frac{{\partial }_2{h}_3}{h_3}$，${\Gamma }_{33}^3=\frac{{\partial }_3{h}_3}{h_3}$

注意： ${\partial }_i={\partial }_{u_i}=\frac{\partial }{\partial u_i}$，( $i=1,2,3$ )。

把(19)式用曲线正交坐标基底 ${\left(\frac{\partial }{\partial u_i}\right)}^a$ 展开为：

$\begin{array}{l}{g}^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e=C_{11}{\left(\frac{\partial }{\partial u_1}\right)}^a{\left(\frac{\partial }{\partial u_1}\right)}^b+C_{12}{\left(\frac{\partial }{\partial u_1}\right)}^a{\left(\frac{\partial }{\partial u_2}\right)}^b+C_{13}{\left(\frac{\partial }{\partial u_1}\right)}^a{\left(\frac{\partial }{\partial u_3}\right)}^b\\ +C_{21}{\left(\frac{\partial }{\partial u_2}\right)}^a{\left(\frac{\partial }{\partial u_1}\right)}^b+C_{22}{\left(\frac{\partial }{\partial u_2}\right)}^a{\left(\frac{\partial }{\partial u_2}\right)}^b+C_{23}{\left(\frac{\partial }{\partial u_2}\right)}^a{\left(\frac{\partial }{\partial u_3}\right)}^b\\ +C_{31}{\left(\frac{\partial }{\partial u_3}\right)}^a{\left(\frac{\partial }{\partial u_1}\right)}^b+C_{32}{\left(\frac{\partial }{\partial u_3}\right)}^a{\left(\frac{\partial }{\partial u_2}\right)}^b+C_{33}{\left(\frac{\partial }{\partial u_3}\right)}^a{\left(\frac{\partial }{\partial u_3}\right)}^b\end{array}$ (22)

其中：

$C_{11}=\left(g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e\right){\left(\text{d}{u}_1\right)}_a{\left(\text{d}{u}_1\right)}_b$，$C_{12}=\left(g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e\right){\left(\text{d}{u}_1\right)}_a{\left(\text{d}{u}_2\right)}_b$

$C_{13}=\left(g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e\right){\left(\text{d}{u}_1\right)}_a{\left(\text{d}{u}_3\right)}_b$，$C_{21}=\left(g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e\right){\left(\text{d}{u}_2\right)}_a{\left(\text{d}{u}_1\right)}_b$

$C_{22}=\left(g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e\right){\left(\text{d}{u}_2\right)}_a{\left(\text{d}{u}_2\right)}_b$，$C_{23}=\left(g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e\right){\left(\text{d}{u}_2\right)}_a{\left(\text{d}{u}_3\right)}_b$ (23)

$C_{31}=\left(g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e\right){\left(\text{d}{u}_3\right)}_a{\left(\text{d}{u}_1\right)}_b$，$C_{32}=\left(g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e\right){\left(\text{d}{u}_3\right)}_a{\left(\text{d}{u}_2\right)}_b$

$C_{33}=\left(g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e\right){\left(\text{d}{u}_3\right)}_a{\left(\text{d}{u}_3\right)}_b$

其中 ${\left(\text{d}{u}_i\right)}_a$ 是 ${\left(\frac{\partial }{\partial u_i}\right)}^a$ 的对偶坐标基底。

利用正交归一基底和坐标基底的关系(17)式，则(19)式用曲线正交坐标系的正交归一基底展开为：

$\begin{array}{l}{g}^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e=C_{11}{h}_1{h}_1{\left(e_1\right)}^a{\left(e_1\right)}^b+C_{12}{h}_1{h}_2{\left(e_1\right)}^a{\left(e_2\right)}^b+C_{13}{h}_1{h}_3{\left(e_1\right)}^a{\left(e_3\right)}^b\\ +C_{21}{h}_2{h}_1{\left(e_2\right)}^a{\left(e_1\right)}^b+C_{22}{h}_2{h}_2{\left(e_2\right)}^a{\left(e_2\right)}^b+C_{23}{h}_2{h}_3{\left(e_2\right)}^a{\left(e_3\right)}^b\\ +C_{31}{h}_3{h}_1{\left(e_3\right)}^a{\left(e_1\right)}^b+C_{32}{h}_3{h}_2{\left(e_3\right)}^a{\left(e_2\right)}^b+C_{33}{h}_3{h}_3{\left(e_3\right)}^a{\left(e_3\right)}^b\end{array}$ (24)

我们这里计算(24)式右边第一项。(23)式第一项为：

$\begin{array}{c}{C}_{11}=\left(g^{ac}{\partial }_c{A}^b+g^{ac}{\Gamma }_{ce}^b{A}^e\right){\left(\text{d}{u}_1\right)}_a{\left(\text{d}{u}_1\right)}_b=g^{1i}{\partial }_i{A}^{u_1}+g^{1i}{\Gamma }_{ie}^1{A}^e=g^{11}{\partial }_1{A}^{u_1}+g^{11}{\Gamma }_{1e}^1{A}^e\\ =g^{11}{\partial }_1{A}^{u_1}+g^{11}{\Gamma }_{11}^1{A}^{u_1}+g^{11}{\Gamma }_{12}^1{A}^{u_2}+g^{11}{\Gamma }_{13}^1{A}^{u_3}=g^{11}{\partial }_1\left(\frac{A^1}{h_1}\right)+g^{11}{\Gamma }_{11}^1\frac{A^1}{h_1}+g^{11}{\Gamma }_{12}^1\frac{A^2}{h_2}+g^{11}{\Gamma }_{13}^1\frac{A^3}{h_3}\end{array}$ (25)

上面最后一个等号我们利用了关系(18)。则得到：

$C_{11}{h}_1{h}_1=\frac{1}{h_1}{\partial }_1{A}^1+\frac{1}{h_1{h}_2}\left({\partial }_2{h}_1\right)A^2+\frac{1}{h_1{h}_3}\left({\partial }_3{h}_1\right)A^3$ (26)

(24)式中其它的系数 $C_{12}{h}_1{h}_2\cdot \cdot \cdot$ 可用类似的计算得到。最后把所求得的系数带入(24)式得到：

$\begin{array}{c}\stackrel{\to }{\nabla }\stackrel{\to }{A}=\left[\frac{1}{h_1}{\partial }_1{A}^1+\frac{1}{h_1{h}_2}\left({\partial }_2{h}_1\right)A^2+\frac{1}{h_1{h}_3}\left({\partial }_3{h}_1\right)A^3\right]{\stackrel{\to }{e}}_1{\stackrel{\to }{e}}_1+\left[\frac{1}{h_1}{\partial }_1{A}^2-\frac{1}{h_1{h}_2}\left({\partial }_2{h}_1\right)A^1\right]{\stackrel{\to }{e}}_1{\stackrel{\to }{e}}_2\\ +\left[\frac{1}{h_1}{\partial }_1{A}^3-\frac{1}{h_1{h}_3}\left({\partial }_3{h}_1\right)A^1\right]{\stackrel{\to }{e}}_1{\stackrel{\to }{e}}_3+\left[\frac{1}{h_2}{\partial }_2{A}^1-\frac{1}{h_1{h}_2}\left({\partial }_1{h}_2\right)A^2\right]{\stackrel{\to }{e}}_2{\stackrel{\to }{e}}_1\\ +\left[\frac{1}{h_2}{\partial }_2{A}^2+\frac{1}{h_1{h}_2}\left({\partial }_1{h}_2\right)A^1+\frac{1}{h_2{h}_3}\left({\partial }_3{h}_2\right)A^3\right]{\stackrel{\to }{e}}_2{\stackrel{\to }{e}}_2\\ +\left[\frac{1}{h_2}{\partial }_2{A}^3-\frac{1}{h_2{h}_3}\left({\partial }_3{h}_2\right)A^2\right]{\stackrel{\to }{e}}_2{\stackrel{\to }{e}}_3+\left[\frac{1}{h_3}{\partial }_3{A}^1-\frac{1}{h_1{h}_3}\left({\partial }_1{h}_3\right)A^3\right]{\stackrel{\to }{e}}_3{\stackrel{\to }{e}}_1\\ +\left[\frac{1}{h_3}{\partial }_3{A}^2-\frac{1}{h_2{h}_3}\left({\partial }_2{h}_3\right)A^3\right]{\stackrel{\to }{e}}_3{\stackrel{\to }{e}}_2+\left[\frac{1}{h_3}{\partial }_3{A}^3+\frac{1}{h_1{h}_3}\left({\partial }_1{h}_3\right)A^1+\frac{1}{h_3{h}_2}\left({\partial }_2{h}_3\right)A^2\right]{\stackrel{\to }{e}}_3{\stackrel{\to }{e}}_3\end{array}$ (27)

这就是 $\stackrel{\to }{\nabla }\stackrel{\to }{A}$ 在曲线正交坐标系的一般表达式，目前我们在通常的教科书中没有找到这个结果。常见的曲线正交坐标系有直角坐标系、柱坐标系和球坐标系，我们以这三个例子为例，得出具体的表达式。

1) 直角坐标系下

$u_1=x,u_2=y,u_3=z$

$h_1=1,h_2=1,h_3=1$

$\begin{array}{l}\stackrel{\to }{\nabla }\stackrel{\to }{A}={\partial }_x{A}^1{\stackrel{\to }{e}}_x{\stackrel{\to }{e}}_x+{\partial }_x{A}^2{\stackrel{\to }{e}}_x{\stackrel{\to }{e}}_y+{\partial }_x{A}^3{\stackrel{\to }{e}}_x{\stackrel{\to }{e}}_z\\ +{\partial }_y{A}^1{\stackrel{\to }{e}}_y{\stackrel{\to }{e}}_x+{\partial }_y{A}^2{\stackrel{\to }{e}}_y{\stackrel{\to }{e}}_y+{\partial }_y{A}^3{\stackrel{\to }{e}}_y{\stackrel{\to }{e}}_z\\ +{\partial }_z{A}^1{\stackrel{\to }{e}}_z{\stackrel{\to }{e}}_x+{\partial }_z{A}^2{\stackrel{\to }{e}}_z{\stackrel{\to }{e}}_y+{\partial }_z{A}^3{\stackrel{\to }{e}}_z{\stackrel{\to }{e}}_z\end{array}$ (28)

2) 柱坐标系下

$u_1=r,u_2=\varphi ,u_3=z$

利用变换关系

$x=r\cos \varphi ,y=r\sin \varphi ,z=z$

带入(12)式可算出：

$h_1=1,h_2=r,h_3=1$

则得到：

$\begin{array}{l}\stackrel{\to }{\nabla }\stackrel{\to }{A}={\partial }_r{A}^1{\stackrel{\to }{e}}_r{\stackrel{\to }{e}}_r+{\partial }_r{A}^2{\stackrel{\to }{e}}_r{\stackrel{\to }{e}}_{\varphi }+{\partial }_r{A}^3{\stackrel{\to }{e}}_r{\stackrel{\to }{e}}_z\\ +\left[\frac{1}{r}{\partial }_{\varphi }{A}^1-\frac{1}{r}{A}^2\right]{\stackrel{\to }{e}}_{\varphi }{\stackrel{\to }{e}}_r+\left[\frac{1}{r}{\partial }_{\varphi }{A}^2+\frac{1}{r}{A}^1\right]{\stackrel{\to }{e}}_{\varphi }{\stackrel{\to }{e}}_{\varphi }+\frac{1}{r}{\partial }_2{A}^3{\stackrel{\to }{e}}_{\varphi }{\stackrel{\to }{e}}_z\\ +{\partial }_z{A}^1{\stackrel{\to }{e}}_z{\stackrel{\to }{e}}_r+{\partial }_z{A}^2{\stackrel{\to }{e}}_z{\stackrel{\to }{e}}_{\varphi }+{\partial }_z{A}^3{\stackrel{\to }{e}}_z{\stackrel{\to }{e}}_z\end{array}$ (29)

3) 球坐标系下

$u_1=r,u_2=\theta ,u_3=\varphi$

$h_1=1,h_2=r,h_3=r\sin \theta$

$\begin{array}{l}\stackrel{\to }{\nabla }\stackrel{\to }{A}={\partial }_r{A}^1{\stackrel{\to }{e}}_r{\stackrel{\to }{e}}_r+{\partial }_r{A}^2{\stackrel{\to }{e}}_r{\stackrel{\to }{e}}_{\theta }+{\partial }_r{A}^3{\stackrel{\to }{e}}_r{\stackrel{\to }{e}}_{\varphi }\\ +\left[\frac{1}{r}{\partial }_{\theta }{A}^1-\frac{1}{r}{A}^2\right]{\stackrel{\to }{e}}_{\theta }{\stackrel{\to }{e}}_r+\left[\frac{1}{r}{\partial }_{\theta }{A}^2+\frac{1}{r}{A}^1\right]{\stackrel{\to }{e}}_{\theta }{\stackrel{\to }{e}}_{\theta }+\frac{1}{r}{\partial }_{\theta }{A}^3{\stackrel{\to }{e}}_{\theta }{\stackrel{\to }{e}}_{\varphi }\\ +\left[\frac{1}{r\sin \theta }{\partial }_{\varphi }{A}^1-\frac{1}{r}{A}^3\right]{\stackrel{\to }{e}}_{\varphi }{\stackrel{\to }{e}}_r\\ +\left[\frac{1}{r\sin \theta }{\partial }_{\varphi }{A}^2-\frac{\cos \theta }{r\sin \theta }{A}^3\right]{\stackrel{\to }{e}}_{\varphi }{\stackrel{\to }{e}}_{\theta }\\ +\left[\frac{1}{r\sin \theta }{\partial }_{\varphi }{A}^3+\frac{1}{r}{A}^1+\frac{\cos \theta }{r\sin \theta }{A}^2\right]{\stackrel{\to }{e}}_{\varphi }{\stackrel{\to }{e}}_{\varphi }\end{array}$ (30)

4. 计算 ${\nabla }^2\stackrel{\to }{A}$

${\nabla }^2\stackrel{\to }{A}$ 用抽象指标可表示为 ${\nabla }_a{\nabla }^a{A}^b$，则：

${\nabla }^2\stackrel{\to }{A}={\nabla }_a{\nabla }^a{A}^b=g^{ca}{\nabla }_c{\nabla }_a{A}^b$ (31)

$=g^{ca}\left[{\partial }_c\left({\nabla }_a{A}^b\right)+{\Gamma }_{cd}^b{\nabla }_a{A}^d-{\Gamma }_{ca}^d{\nabla }_d{A}^d\right]$

$=g^{ca}\left\{{\partial }_c\left({\partial }_a{A}^b+{\Gamma }_{ad}^b{A}^d\right)+{\Gamma }_{cd}^b\left({\partial }_a{A}^d+{\Gamma }_{ae}^d{A}^e\right)-{\Gamma }_{ca}^d\left({\partial }_d{A}^b+{\Gamma }_{de}^b{A}^e\right)\right\}$

$\begin{array}{l}=g^{ca}{\partial }_c{\partial }_a{A}^b+g^{ca}{\partial }_c\left({\Gamma }_{ad}^b\right)A^d+g^{ca}{\Gamma }_{ad}^b{\partial }_c{A}^d+g^{ca}{\Gamma }_{cd}^b{\partial }_a{A}^d\\ +g^{ca}{\Gamma }_{cd}^b{\Gamma }_{ae}^d{A}^e-g^{ca}{\Gamma }_{ca}^d{\partial }_d{A}^b-g^{ca}{\Gamma }_{ca}^d{\Gamma }_{de}^b{A}^e\end{array}$ (32)

(31)式用曲线正交坐标基底 ${\left(\frac{\partial }{\partial u_i}\right)}^a$ 展开为：

${\nabla }^2\stackrel{\to }{A}=g^{ca}{\nabla }_c{\nabla }_a{A}^b=C_1{\left(\frac{\partial }{\partial u_1}\right)}^b+C_2{\left(\frac{\partial }{\partial u_2}\right)}^b+C_3{\left(\frac{\partial }{\partial u_3}\right)}^b$ (33)

其中：

$\begin{array}{l}{C}_1=g^{ca}{\nabla }_c{\nabla }_a{A}^b{\left(\text{d}{u}_1\right)}_b\\ C_2=g^{ca}{\nabla }_c{\nabla }_a{A}^b{\left(\text{d}{u}_2\right)}_b\\ C_3=g^{ca}{\nabla }_c{\nabla }_a{A}^b{\left(\text{d}{u}_3\right)}_b\end{array}$ (34)

利用正交归一基底和坐标基底的关系(17)式，(31)式可用曲线正交坐标系的正交归一基底展开为：

${\nabla }^2\stackrel{\to }{A}=g^{ca}{\nabla }_c{\nabla }_a{A}^b=C_1{h}_1{\left(e_1\right)}^b+C_2{h}_2{\left(e_2\right)}^b+C_3{h}_3{\left(e_3\right)}^b$ (35)

下面我们求 $C_1{h}_1$。

$\begin{array}{c}{C}_1{h}_1=g^{ca}{\nabla }_c{\nabla }_a{A}^b{\left(\text{d}{u}_1\right)}_b\\ =h_1{g}^{ca}{\partial }_c{\partial }_a{A}^b{\left(\text{d}{u}_1\right)}_b+h_1{g}^{ca}{\partial }_c\left({\Gamma }_{ad}^b\right)A^d{\left(\text{d}{u}_1\right)}_b+h_1{g}^{ca}{\Gamma }_{ad}^b{\partial }_c{A}^d{\left(\text{d}{u}_1\right)}_b\\ +h_1{g}^{ca}{\Gamma }_{cd}^b{\partial }_a{A}^d{\left(\text{d}{u}_1\right)}_b+h_1{g}^{ca}{\Gamma }_{cd}^b{\Gamma }_{ae}^d{A}^e{\left(\text{d}{u}_1\right)}_b-h_1{g}^{ca}{\Gamma }_{ca}^d{\partial }_d{A}^b{\left(\text{d}{u}_1\right)}_b\\ -h_1{g}^{ca}{\Gamma }_{ca}^d{\Gamma }_{de}^b{A}^e{\left(\text{d}{u}_1\right)}_b\end{array}$ (36)

上式第二个等号用了(32)式。我们以(36)式第二个等号后的第二项的计算为例。

$\begin{array}{l}{h}_1{g}^{ca}{\partial }_c\left({\Gamma }_{ad}^b\right)A^d{\left(\text{d}{u}_1\right)}_b=h_1{g}^{ij}{\partial }_i\left({\Gamma }_{jd}^1\right)A^d\\ =h_1{g}^{11}{\partial }_1\left({\Gamma }_{1d}^1\right)A^d+h_1{g}^{22}{\partial }_2\left({\Gamma }_{2d}^1\right)A^d+h_1{g}^{33}{\partial }_3\left({\Gamma }_{3d}^1\right)A^d\\ =h_1{g}^{11}{\partial }_1\left({\Gamma }_{11}^1\right)\frac{A^1}{h_1}+h_1{g}^{11}{\partial }_1\left({\Gamma }_{12}^1\right)\frac{A^2}{h_2}+h_1{g}^{11}{\partial }_1\left({\Gamma }_{13}^1\right)\frac{A^3}{h_3}\\ +h_1{g}^{22}{\partial }_2\left({\Gamma }_{21}^1\right)\frac{A^1}{h_1}+h_1{g}^{22}{\partial }_2\left({\Gamma }_{22}^1\right)\frac{A^2}{h_2}+h_1{g}^{33}{\partial }_2\left({\Gamma }_{23}^1\right)\frac{A^3}{h_3}\\ +h_1{g}^{33}{\partial }_3\left({\Gamma }_{31}^1\right)\frac{A^1}{h_1}+h_1{g}^{33}{\partial }_3\left({\Gamma }_{32}^1\right)\frac{A^2}{h_2}+h_1{g}^{33}{\partial }_3\left({\Gamma }_{33}^1\right)\frac{A^3}{h_3}\end{array}$ (37)

同样的办法求出(36)式中剩下的6项，即可求得 (35)式中 ${\left(e_1\right)}^b$ 前的系数：

$\begin{array}{c}{C}_1{h}_1=A^1\left(-1\right)\left\{\frac{{\left({\partial }_3{h}_1\right)}^2}{h_1^2{h}_3^2}+\frac{{\left({\partial }_2{h}_1\right)}^2}{h_1^2{h}_2^2}+\frac{{\left({\partial }_1{h}_2\right)}^2}{h_1^2{h}_2^2}+\frac{{\left({\partial }_1{h}_3\right)}^2}{h_1^2{h}_3^2}\right\}\\ +{\partial }_1{A}^1\left[-\frac{{\partial }_1{h}_1}{h_1^3}+\frac{{\partial }_1{h}_2}{h_1^2{h}_2}+\frac{{\partial }_1{h}_3}{h_1^2{h}_3}\right]+{\partial }_2{A}^1\left[\frac{{\partial }_2{h}_1}{h_1{h}_2^2}-\frac{{\partial }_2{h}_2}{h_2^3}+\frac{{\partial }_2{h}_3}{h_2^2{h}_3}\right]\\ +{\partial }_3{A}^1\left[\frac{{\partial }_3{h}_1}{h_1{h}_3^2}-\frac{{\partial }_3{h}_3}{h_3^3}+\frac{{\partial }_3{h}_2}{h_2{h}_3^2}\right]+{\partial }_1{\partial }_1{A}^1\frac{1}{h_1^2}+{\partial }_2{\partial }_2{A}^1\frac{1}{h_2^2}+{\partial }_3{\partial }_3{A}^1\frac{1}{h_3^2}\\ +A^2\left[-\frac{{\partial }_2{h}_1{\partial }_1{h}_1}{h_1^3{h}_2}+\frac{{\partial }_2{h}_2{\partial }_1{h}_2}{h_1{h}_2^3}-\frac{{\partial }_2{h}_3{\partial }_1{h}_2}{h_1{h}_2^2{h}_3}+\frac{{\partial }_2{h}_1{\partial }_1{h}_3}{h_1^2{h}_2{h}_3}-\frac{{\partial }_2{h}_3{\partial }_1{h}_3}{h_1{h}_2{h}_3^2}+\frac{{\partial }_1{\partial }_2{h}_1}{h_1^2{h}_2}-\frac{{\partial }_1{\partial }_2{h}_2}{h_1{h}_2^2}\right]\\ +{\partial }_1{A}^2\frac{2{\partial }_2{h}_1}{h_1^2{h}_2}-{\partial }_2{A}^2\frac{2{\partial }_1{h}_2}{h_1{h}_2^2}+{\partial }_1{A}^3\frac{2{\partial }_3{h}_1}{h_1^2{h}_3}-{\partial }_3{A}^3\frac{2{\partial }_1{h}_3}{h_1{h}_3^2}\\ +A^3\left[-\frac{{\partial }_3{h}_1{\partial }_1{h}_1}{h_1^3{h}_3}+\frac{{\partial }_3{h}_1{\partial }_1{h}_2}{h_1^2{h}_2{h}_3}-\frac{{\partial }_3{h}_2{\partial }_1{h}_2}{h_1{h}_2^2{h}_3}-\frac{{\partial }_3{h}_2{\partial }_1{h}_3}{h_1{h}_2{h}_3^2}+\frac{{\partial }_3{h}_3{\partial }_1{h}_3}{h_1{h}_3^3}+\frac{{\partial }_1{\partial }_3{h}_1}{h_1^2{h}_3}-\frac{{\partial }_1{\partial }_3{h}_3}{h_1{h}_3^2}\right]\end{array}$ (38)

根据对称性，把(38)式利用指标轮换( $1\to 2,\text{2}\to 3,\text{3}\to 1$ )即可直接写出 ${\left(e_2\right)}^b$ 前的系数：

$\begin{array}{c}{C}_2{h}_2=A^2\left(-1\right)\left\{\frac{{\left({\partial }_1{h}_2\right)}^2}{h_2^2{h}_1^2}+\frac{{\left({\partial }_3{h}_2\right)}^2}{h_2^2{h}_3^2}+\frac{{\left({\partial }_2{h}_3\right)}^2}{h_2^2{h}_3^2}+\frac{{\left({\partial }_2{h}_1\right)}^2}{h_2^2{h}_1^2}\right\}\\ +{\partial }_2{A}^2\left[-\frac{{\partial }_2{h}_2}{h_2^3}+\frac{{\partial }_2{h}_3}{h_2^2{h}_3}+\frac{{\partial }_2{h}_1}{h_2^2{h}_1}\right]+{\partial }_3{A}^2\left[\frac{{\partial }_3{h}_2}{h_2{h}_3^2}-\frac{{\partial }_3{h}_3}{h_3^3}+\frac{{\partial }_3{h}_1}{h_3^2{h}_1}\right]\\ +{\partial }_1{A}^2\left[\frac{{\partial }_1{h}_2}{h_2{h}_1^2}-\frac{{\partial }_1{h}_1}{h_1^3}+\frac{{\partial }_1{h}_3}{h_3{h}_1^2}\right]+{\partial }_2{\partial }_2{A}^2\frac{1}{h_2^2}+{\partial }_3{\partial }_3{A}^2\frac{1}{h_3^2}+{\partial }_1{\partial }_1{A}^2\frac{1}{h_1^2}\\ +A^3\left[-\frac{{\partial }_3{h}_2{\partial }_2{h}_2}{h_2^3{h}_3}+\frac{{\partial }_3{h}_3{\partial }_2{h}_3}{h_2{h}_3^3}-\frac{{\partial }_3{h}_1{\partial }_2{h}_3}{h_2{h}_3^2{h}_1}+\frac{{\partial }_3{h}_2{\partial }_2{h}_1}{h_2^2{h}_3{h}_1}-\frac{{\partial }_3{h}_1{\partial }_2{h}_1}{h_2{h}_3{h}_1^2}+\frac{{\partial }_2{\partial }_3{h}_2}{h_2^2{h}_3}-\frac{{\partial }_2{\partial }_3{h}_3}{h_2{h}_3^2}\right]\\ +{\partial }_2{A}^3\frac{2{\partial }_3{h}_2}{h_2^2{h}_3}-{\partial }_3{A}^3\frac{2{\partial }_2{h}_3}{h_2{h}_3^2}+{\partial }_2{A}^1\frac{2{\partial }_1{h}_2}{h_2^2{h}_1}-{\partial }_1{A}^1\frac{2{\partial }_2{h}_1}{h_2{h}_1^2}\\ +A^1\left[-\frac{{\partial }_1{h}_2{\partial }_2{h}_2}{h_2^3{h}_1}+\frac{{\partial }_1{h}_2{\partial }_2{h}_3}{h_2^2{h}_3{h}_1}-\frac{{\partial }_1{h}_3{\partial }_2{h}_3}{h_2{h}_3^2{h}_1}-\frac{{\partial }_1{h}_3{\partial }_2{h}_1}{h_2{h}_3{h}_1^2}+\frac{{\partial }_1{h}_1{\partial }_2{h}_1}{h_2{h}_1^3}+\frac{{\partial }_2{\partial }_1{h}_2}{h_2^2{h}_1}-\frac{{\partial }_2{\partial }_1{h}_1}{h_2{h}_1^2}\right]\end{array}$ (39)