附录
\(\vec{r}/r^3\) 形式矢量场的曲线积分¶
如上图所示，矢量场 \(\vec{F}\) 在点 \(P\) 处为 \(\displaystyle \vec{F} = k \frac{\vec{r}}{r^3}\)，其中，\(\vec{r}\) 为点 \(P\) 相对于原点 \(O\) 的位置矢量，\(r = \abs{\vec{r}\tm}\)，\(k\) 为常数
沿着路径 \(L\) 从点 \(A\) 到点 \(B\)，矢量场 \(\vec{F}\) 的曲线积分
\[
\int_L \vec{F} \cdot \d{\vec{r}}
= k \int_L \frac{\vec{r} \cdot \d{\vec{r}}}{r^3}
\tag{1} \label{eq:curve-integral}
\]
其中，根据对标量积求微分的法则
\[
\d{\rb{\vec{r} \cdot \vec{r}\tm}} = \vec{r} \cdot \d{\vec{r}} + \d{\vec{r}} \cdot \vec{r}
\ \Rightarrow\
\vec{r} \cdot \d{\vec{r}} = \frac{1}{2} \d{\rb{\vec{r} \cdot \vec{r}\tm}}
= \frac{1}{2} \d{\rb{r^2\tm }} = r \,\d{r}
\]
\[
\begin{align}
& \d{\rb{\vec{r} \cdot \vec{r}\tm}} = \vec{r} \cdot \d{\vec{r}} + \d{\vec{r}} \cdot \vec{r} \\
& \Rightarrow\
\vec{r} \cdot \d{\vec{r}} = \frac{1}{2} \d{\rb{\vec{r} \cdot \vec{r}\tm}}
= \frac{1}{2} \d{\rb{r^2\tm }} = r \,\d{r}
\end{align}
\]
直角坐标系中推导上式
在直角坐标系中
\[
\vec{r} = x \,\vec{i} + y \,\vec{j} + z \,\vec{k},\
\d{\vec{r}} = \d{x} \,\vec{i} + \d{y} \,\vec{j} + \d{z} \,\vec{k}
\]
因此
\[
\begin{align}
& \vec{r} \cdot \d{\vec{r}}
= x \,\d{x} + y \,\d{y} + z \,\d{z}
= \frac{1}{2} \d{x^2} + \frac{1}{2} \d{y^2} + \frac{1}{2} \d{z^2} \\
& = \frac{1}{2} \d{\rb{x^2 + y^2 + z^2}}
= \frac{1}{2} \d{\rb{r^2}}
= r \,\d{r}
\end{align}
\]
\[
\begin{align}
& \vec{r} \cdot \d{\vec{r}}
= x \,\d{x} + y \,\d{y} + z \,\d{z} \\
& = \frac{1}{2} \d{x^2} + \frac{1}{2} \d{y^2} + \frac{1}{2} \d{z^2} \\
& = \frac{1}{2} \d{\rb{x^2 + y^2 + z^2}} \\
& = \frac{1}{2} \d{\rb{r^2}}
= r \,\d{r}
\end{align}
\]
将上式的结果代入式 \(\eqref{eq:curve-integral}\)，原先对 \(L\) 的曲线积分就转变为了对距离 \(r\) 的一元函数积分
\[
k \int_L \frac{\vec{r} \cdot \d{\vec{r}}}{r^3}
= k \int_{r_A}^{r_B} \frac{r \,\d{r}}{r^3}
= \left. \frac{-k}{r} \right|_{r_A}^{r_B}
= k \rb{\frac{1}{r_A} - \frac{1}{r_B}}
\]
\[
\begin{align}
& k \int_L \frac{\vec{r} \cdot \d{\vec{r}}}{r^3}
= k \int_{r_A}^{r_B} \frac{r \,\d{r}}{r^3}
= \left. \frac{-k}{r} \right|_{r_A}^{r_B} \\
& = k \rb{\frac{1}{r_A} - \frac{1}{r_B}}
\end{align}
\]
上述结果表明，矢量场 \(\vec{F}\) 的曲线积分 \(\eqref{eq:curve-integral}\) 与 \(A\)、\(B\) 两点之间路径 \(L\) 的选择无关，只取决于它们到原点的距离
若一个矢量场的曲线积分结果与路径选择无关，只与始末位置有关，则该矢量场被称为“保守场”，显然，本例中的 \(\vec{F}\) 就是一个保守场
\(1/\abs{\vec{r}-\vec{q}\tm}\) 梯度的计算¶
在直角坐标系中
\[
\vec{r} = x \,\vec{i} + y \,\vec{j} + z \,\vec{k},\
\vec{q} = x' \,\vec{i} + y' \,\vec{j} + z' \,\vec{k}
\]
则
\[
\frac{1}{\abs{\vec{r}-\vec{q}\tm}} = \frac{1}{\sqrt{\rb{x-x'}^2 + \rb{y-y'}^2 + \rb{z-z'}^2}}
\]
考虑只对 \(\vec{r}\) 中的空间变量求导的梯度算符 \(\nabla\)
\[
\nabla f = \frac{\partial f}{\partial x} \,\vec{i} + \frac{\partial f}{\partial y} \,\vec{j} + \frac{\partial f}{\partial z} \,\vec{k}
\]
将其中的 \(f\) 用 \(\displaystyle \frac{1}{\abs{\vec{r}-\vec{q}\tm}}\) 替换，得到
\[
\frac{\partial}{\partial x} \frac{1}{\abs{\vec{r}-\vec{q}\tm}}
= -\frac{1}{2} \frac{2\rb{x-x'}}{\sb{\rb{x-x'}^2 + \rb{y-y'}^2 + \rb{z-z'}^2}^{3/2}}
= - \frac{\rb{x-x'}}{\abs{\vec{r}-\vec{q}\tm}^3}
\]
\[
\begin{align}
& \frac{\partial}{\partial x} \frac{1}{\abs{\vec{r}-\vec{q}\tm}} \\
& = -\frac{1}{2} \frac{2\rb{x-x'}}{\sb{\rb{x-x'}^2 + \rb{y-y'}^2 + \rb{z-z'}^2}^{3/2}} \\
& = - \frac{\rb{x-x'}}{\abs{\vec{r}-\vec{q}\tm}^3}
\end{align}
\]
类似的
\[
\frac{\partial}{\partial y} \frac{1}{\abs{\vec{r}-\vec{q}\tm}} = - \frac{\rb{y-y'}}{\abs{\vec{r}-\vec{q}\tm}^3},\
\frac{\partial}{\partial z} \frac{1}{\abs{\vec{r}-\vec{q}\tm}} = - \frac{\rb{z-z'}}{\abs{\vec{r}-\vec{q}\tm}^3}
\]
\[
\begin{gather}
\frac{\partial}{\partial y} \frac{1}{\abs{\vec{r}-\vec{q}\tm}} = - \frac{\rb{y-y'}}{\abs{\vec{r}-\vec{q}\tm}^3} \\
\frac{\partial}{\partial z} \frac{1}{\abs{\vec{r}-\vec{q}\tm}} = - \frac{\rb{z-z'}}{\abs{\vec{r}-\vec{q}\tm}^3}
\end{gather}
\]
因此
\[
\nabla \frac{1}{\abs{\vec{r}-\vec{q}\tm}}
= - \frac{\rb{x-x'}\,\vec{i} + \rb{y-y'}\,\vec{j} + \rb{z-z'}\,\vec{k}}{\abs{\vec{r}-\vec{q}\tm}^3}
= \hlt{
- \frac{\vec{r}-\vec{q}\tm}{\abs{\vec{r}-\vec{q}\tm}^3}
}
\]
\[
\begin{align}
& \nabla \frac{1}{\abs{\vec{r}-\vec{q}\tm}} \\
& = - \frac{\rb{x-x'}\,\vec{i} + \rb{y-y'}\,\vec{j} + \rb{z-z'}\,\vec{k}}{\abs{\vec{r}-\vec{q}\tm}^3} \\
& = \hlt{
- \frac{\vec{r}-\vec{q}\tm}{\abs{\vec{r}-\vec{q}\tm}^3}
}
\end{align}
\]