微分形式
2026/03/10 10:00
线分布：\(d \vec{l} = (d x_1, d x_2, d x_3), \, d x_i = \frac{d x_i}{d t} dt\)
\[ dl = |d \vec{l}| = \sqrt{\sum_i^3 \Big(\frac{d x_i}{d t}\Big)^2} dt \]
三维空间中确定一条曲线，需要两个方程：
\[ \left \{ \begin{aligned} & f(\vec{x}) = 0 \\ & g(\vec{x}) = 0 \end{aligned} \right. \]
\(\vec{x}(t)\) 在曲线上，\(\vec{x} + d \vec{x}\) 也在曲线上：
\[ \left. \begin{aligned} \underset{= 0}{\underbrace{f(\vec{x} + d \vec{x})}} = \underset{= 0}{\underbrace{f(\vec{x})}} + \underset{= 0}{\underbrace{\ldots}} \iff d f(\vec{x}) &= \sum_i^3 \frac{\partial f}{\partial x_i} d x_i = 0 \\ d g(\vec{x}) &= \sum_i^3 \frac{\partial g}{\partial x_i} d x_i = 0 \end{aligned} \right \} \implies \left \{ \begin{aligned} d x_2 &= d x_2 (d x_1) \\ d x_3 &= d x_3 (d x_1) \end{aligned} \right. \]
面分布：\(d S = d x_1 d x_2\). 将 \(x_1\)-\(x_2\) 系逆时针旋转角度 \(\theta\)，得到 \(y_1\)-\(y_2\) 系，则理应有 \(d x_1 d x_2 = d y_1 d y_2\). 但是如果我们这么做：先写出变换关系
\[ \begin{aligned} d x_1 &= \cos \theta d y_1 - \sin \theta d y_2 \\ d x_2 &= \sin \theta d y_1 + \cos \theta d y_2 \end{aligned} \]
若直接将上式代入计算 \(d x_1 d x_2\)，会得到
\[ d x_1 d x_2 = (\cos^2 \theta - \sin^2 \theta) d y_1 d y_2 + \sin \theta \cos \theta (d y_1^2 - d y_2^2) \]
不合理之处：
- 和角度有关！
- \(d y_1^2, d y_2^2\) 是什么鬼！不可能出现在二重积分中！
人为引入规则： \(d x^2 = 0\)
楔积（wedge product）
\(d x_1 d x_2 \to d x_1 \wedge d x_2\)
条件：\(dx \wedge dx = 0\)
将 \(dx\) 替换为 \(dy + dz\)，可推出反对称性：\(dy \wedge dz = - dz \wedge dy\)
引入楔积后，旋转变换的结果就变成了
\[ \begin{aligned} d x_1 \wedge d x_2 &= \cos^2 \theta d y_1 \wedge d y_2 - \sin^2 \theta d y_2 \wedge d y_1 + \sin \theta \cos \theta (d y_1 \wedge d y_1 - d y_2 \wedge d y_2) \\ &= (\cos^2 \theta + \sin^2 \theta) d y_1 \wedge d y_2 \\ &= d y_1 \wedge d y_2 \end{aligned} \]
同理，对于极坐标 \(x_1 = r \cos \theta, \, x_2 = r \sin \theta\)，有
\[ \begin{aligned} d x_1 \wedge d x_2 &= \cos^2 \theta r d r \wedge d \theta - \sin^2 \theta r d r \wedge d \theta + \sin \theta \cos \theta (d r \wedge d r - r^2 d \theta \wedge d \theta) \\ &= r d r \wedge d \theta \end{aligned} \]
曲面：逆着坐标轴方向做投影（为了保持右手系）
\[ d \vec{S} = (d x_2 \wedge d x_3, d x_3 \wedge d x_1, d x_1 \wedge d x_2) \]
球坐标
\[ \left \{ \begin{aligned} x_1 &= r \sin \theta \cos \varphi \\ x_2 &= r \sin \theta \sin \varphi \\ x_3 &= r \cos \theta \end{aligned} \right. \implies \left \{ \begin{aligned} d x_1 &= \sin \theta \cos \varphi d r + r \cos \theta \cos \varphi d \theta - r \sin \theta \sin \varphi d \varphi \\ d x_2 &= \sin \theta \sin \varphi d r + r \cos \theta \sin \varphi d \theta + r \sin \theta \cos \varphi d \varphi \\ d x_3 &= \cos \theta d r - r \sin \theta d \theta \end{aligned} \right. \]
取个特例，在 \(r = 1\) 的球面上，\(dr = 0\)，
\[ d \vec{S} = (\sin^2 \theta \cos \varphi, \sin^2 \theta \sin \varphi, \sin \theta \cos \theta) \, d \theta \wedge d \varphi \]
容易得到 \(|d \vec{S}| = \sin \theta d \theta d \varphi\)，显然正确。
体分布：\(d V = d x_1 \wedge d x_2 \wedge d x_3\)
实际积分的时候怎么用楔积？
直接去掉 \(\wedge\)，交换积分顺序也不用加负号，只要关注上下限就可以了。