DOI:10.12677/IJM.2021.102014,PDF,HTML,XML,被引量下载: 570浏览: 1,221科研立项经费支持
作者:陈葛锐^*,贺梦冬,陈小艳：中南林业科技大学理学院，湖南 长沙
关键词:曲线正交坐标系；微分几何；_∇^→_A^→；∇²_A^→；Curvilinear Orthogonal Coordinates；Differential Geometry；_∇^→_A^→；∇²_A^→

文章引用：陈葛锐, 贺梦冬, 陈小艳. 曲线正交坐标系下 _∇ ^→ _A ^→与∇ ² _A ^→的一种推导[J]. 力学研究, 2021, 10(2): 143-152. https://doi.org/10.12677/IJM.2021.102014

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)

同理把(39)式利用指标轮换( $1\to 2,\text{2}\to 3,\text{3}\to 1$)可直接写出 ${\left(e_3\right)}^b$前的系数：

$\begin{array}{c}{C}_3{h}_3=A^3\left(-1\right)\left\{\frac{{\left({\partial }_2{h}_3\right)}^2}{h_3^2{h}_2^2}+\frac{{\left({\partial }_1{h}_3\right)}^2}{h_3^2{h}_1^2}+\frac{{\left({\partial }_3{h}_1\right)}^2}{h_3^2{h}_1^2}+\frac{{\left({\partial }_3{h}_2\right)}^2}{h_3^2{h}_2^2}\right\}\\ +{\partial }_3{A}^3\left[-\frac{{\partial }_3{h}_3}{h_3^3}+\frac{{\partial }_3{h}_1}{h_3^2{h}_1}+\frac{{\partial }_3{h}_2}{h_3^2{h}_2}\right]+{\partial }_1{A}^3\left[\frac{{\partial }_1{h}_3}{h_3{h}_1^2}-\frac{{\partial }_1{h}_1}{h_1^3}+\frac{{\partial }_1{h}_2}{h_1^2{h}_2}\right]\\ +{\partial }_2{A}^3\left[\frac{{\partial }_2{h}_3}{h_3{h}_2^2}-\frac{{\partial }_2{h}_2}{h_2^3}+\frac{{\partial }_2{h}_1}{h_1{h}_2^2}\right]+{\partial }_3{\partial }_3{A}^3\frac{1}{h_3^2}+{\partial }_1{\partial }_1{A}^3\frac{1}{h_1^2}+{\partial }_2{\partial }_2{A}^3\frac{1}{h_2^2}\\ +A^1\left[-\frac{{\partial }_1{h}_3{\partial }_3{h}_3}{h_3^3{h}_1}+\frac{{\partial }_1{h}_1{\partial }_3{h}_1}{h_3{h}_1^3}-\frac{{\partial }_1{h}_2{\partial }_3{h}_1}{h_3{h}_1^2{h}_2}+\frac{{\partial }_1{h}_3{\partial }_3{h}_2}{h_3^2{h}_1{h}_2}-\frac{{\partial }_1{h}_2{\partial }_3{h}_2}{h_3{h}_1{h}_2^2}+\frac{{\partial }_3{\partial }_1{h}_3}{h_3^2{h}_1}-\frac{{\partial }_3{\partial }_1{h}_1}{h_3{h}_1^2}\right]\\ +{\partial }_3{A}^1\frac{2{\partial }_1{h}_3}{h_3^2{h}_1}-{\partial }_1{A}^1\frac{2{\partial }_3{h}_1}{h_3{h}_1^2}+{\partial }_3{A}^2\frac{2{\partial }_2{h}_3}{h_3^2{h}_2}-{\partial }_2{A}^2\frac{2{\partial }_3{h}_2}{h_3{h}_2^2}\\ +A^2\left[-\frac{{\partial }_2{h}_3{\partial }_3{h}_3}{h_3^3{h}_2}+\frac{{\partial }_2{h}_3{\partial }_3{h}_1}{h_3^2{h}_1{h}_2}-\frac{{\partial }_2{h}_1{\partial }_3{h}_1}{h_3{h}_1^2{h}_2}-\frac{{\partial }_2{h}_1{\partial }_3{h}_2}{h_3{h}_1{h}_2^2}+\frac{{\partial }_2{h}_2{\partial }_3{h}_2}{h_3{h}_2^3}+\frac{{\partial }_3{\partial }_2{h}_3}{h_3^2{h}_2}-\frac{{\partial }_3{\partial }_2{h}_2}{h_3{h}_2^2}\right]\end{array}$(40)

最后把(38)式、(39)式、(40)式带入(35)式就得到了曲线正交坐标系下 ${\nabla }^2\stackrel{\to }{A}$的一般表达式。需要说明的是， ${\nabla }^2\stackrel{\to }{A}$在正交曲线坐标系的一般表达式在题解 [3] 中附录部分不加证明地直接给出。题解 [3] 给出的是相对紧凑的形式，而我们给出的是完全展开的形式，简单计算容易发现两个结果是一样的。

下面我们给出 ${\nabla }^2\stackrel{\to }{A}$在直角坐标系、柱坐标系和球坐标系的具体表达式。

1) 直角坐标系

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

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

${\nabla }^2\stackrel{\to }{A}={\nabla }^2{A}^1{\stackrel{\to }{e}}_x+{\nabla }^2{A}^2{\stackrel{\to }{e}}_y+{\nabla }^2{A}^3{\stackrel{\to }{e}}_z$(41)

2) 柱坐标系

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

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

$\begin{array}{l}{\nabla }^2\stackrel{\to }{A}=\left(-\frac{A^1}{r^2}+{\partial }_1{A}^1\frac{1}{r}+{\partial }_1{\partial }_1{A}^1+{\partial }_2{\partial }_2{A}^1\frac{1}{r^2}+{\partial }_3{\partial }_3{A}^1-{\partial }_2{A}^2\frac{2}{r^2}\right){\stackrel{\to }{e}}_r\\ +\left(\frac{1}{r}{\partial }_1{A}^2+{\partial }_1{\partial }_1{A}^2+{\partial }_2{\partial }_2{A}^2\frac{1}{r^2}+{\partial }_3{\partial }_3{A}^2-\frac{A^2}{r^2}+\frac{2}{r^2}{\partial }_2{A}^1\right){\stackrel{\to }{e}}_{\varphi }\\ +\left(\frac{1}{r}{\partial }_1{A}^3+{\partial }_1{\partial }_1{A}^3+{\partial }_2{\partial }_2{A}^3\frac{1}{r^2}+{\partial }_3{\partial }_3{A}^3\right){\stackrel{\to }{e}}_z\end{array}$(42)

3) 球坐标系下

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

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

$\begin{array}{c}{\nabla }^2\stackrel{\to }{A}=[-\frac{2}{r^2}{A}^1+{\partial }_1{A}^1\frac{2}{r}+{\partial }_2{A}^1\frac{\cos \theta }{r^2\sin \theta }+{\partial }_1{\partial }_1{A}^1+{\partial }_2{\partial }_2{A}^1\frac{1}{r^2}+{\partial }_3{\partial }_3{A}^1\frac{1}{r^2{\left(\sin \theta \right)}^2}\\ -A^2\frac{2\cos \theta }{r^2\sin \theta }-{\partial }_2{A}^2\frac{2}{r^2}-{\partial }_3{A}^3\frac{2}{r^2\sin \theta }]{\stackrel{\to }{e}}_r\\ +[-\frac{1}{r^2{\left(\sin \theta \right)}^2}{A}^2+{\partial }_2{A}^2\frac{\cos \theta }{r^2\sin \theta }+{\partial }_1{A}^2\frac{2}{r}+{\partial }_2{\partial }_2{A}^2\frac{1}{r^2}+{\partial }_3{\partial }_3{A}^2\frac{1}{r^2{\left(\sin \theta \right)}^2}\\ +{\partial }_1{\partial }_1{A}^2-{\partial }_3{A}^3\frac{2\cos \theta }{r^2{\left(\sin \theta \right)}^2}+{\partial }_2{A}^1\frac{2}{r^2}]{\stackrel{\to }{e}}_{\theta }\\ +[-\frac{1}{r^2{\left(\sin \theta \right)}^2}{A}^3+{\partial }_1{A}^3\frac{2}{r}+{\partial }_2{A}^3\frac{\cos \theta }{r^2\sin \theta }+{\partial }_3{\partial }_3{A}^3\frac{1}{r^2{\left(\sin \theta \right)}^2}+{\partial }_1{\partial }_1{A}^3\\ +{\partial }_2{\partial }_2{A}^3\frac{1}{r^2}+{\partial }_3{A}^1\frac{2}{r^2\sin \theta }+{\partial }_3{A}^2\frac{2\cos \theta }{r^2{\left(\sin \theta \right)}^2}]{\stackrel{\to }{e}}_{\varphi }\end{array}$(43)

这样我们就得到了曲线正交坐标系下 $\stackrel{\to }{\nabla }\stackrel{\to }{A}$与 ${\nabla }^2\stackrel{\to }{A}$在直角坐标系、柱坐标系和球坐标系的具体表达式。有兴趣的读者可以照着上面的步骤得出其它曲线正交坐标系下 $\stackrel{\to }{\nabla }\stackrel{\to }{A}$与 ${\nabla }^2\stackrel{\to }{A}$的具体表达式。下面我们给出椭球坐标系与直角坐标系的变换关系，

$\begin{array}{l}x=\sqrt{r^2+a^2}\sin \theta \cos \varphi ,\\ y=\sqrt{r^2+a^2}\sin \theta \sin \varphi ,\\ z=r\cos \theta \end{array}$(44)

请读者朋友自己完成。

5. 结论

本文从微分几何的角度推导了曲线正交坐标系下 $\stackrel{\to }{\nabla }\stackrel{\to }{A}$与 ${\nabla }^2\stackrel{\to }{A}$的一般表达式，以及它们在三维直角坐标系、柱坐标系和球坐标系的具体表达式，这些表达式在相对复杂的物理问题中会涉及到。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，因此在本科物理教学阶段适当指导学生了解一些微分几何知识对学生未来的发展是极有帮助的。文中详细展示了推导计算的每一步过程，这样有助于读者初步了解微分几何的基本思想和方法。

基金项目

湖南省教育厅一般项目19C1895，中南林业科技大学校级教改项目。

NOTES

¹罗杰·彭罗斯(Roger Penrose，1931年8月8日-)，2020年诺贝尔物理学奖获得者，英国数学物理学家、牛津大学数学系名誉教授，他在数学物理方面的工作拥有高度评价，特别是对广义相对论与宇宙学方面的贡献。

²文中的计算均用符号计算软件Mathematica计算。