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微分形式(Pt.II)

2026/03/12 8:00

  • 静电场的环量与通量
  • 微分形式(续)
  • 梯度场
  • 张量计算
\[ \text{电荷} \underset{\text{环量、通量}}{\overset{\text{库仑定律}}{\rightleftharpoons}} \vec{E}(\vec{x}) \]

环量

\[ \int_L \vec{E}(\vec{x}) \cdot d \vec{l} = \phi_{\text{start}}(\vec{x}) - \phi_{\text{end}}(\vec{x}) \implies \oint_L \vec{E}(\vec{x}) \cdot d \vec{l} = 0 \]

静电势

\[ \phi(\vec{x}) = \frac{1}{4 \pi \epsilon_0} \frac{Q}{|\vec{x} - \vec{x}'|} \]

重要关系式(全微分)

\[ \mathrm{d} \frac{1}{|\vec{x} - \vec{x}'|} = - \frac{\vec{x} - \vec{x}'}{|\vec{x} - \vec{x}'|^3} \cdot d \vec{x} \]

通量

\[ \int_S \vec{E}(\vec{x}) \cdot d \vec{S} = \frac{Q}{\varepsilon_0} \]

引出电场线

  1. 电场线不能闭合

    沿着电场线积分有 \(\int \vec{E} \cdot d\vec{l} = \int |\vec{E}| \cdot |d \vec{l}| > 0\),如果有闭合的电场线,则 \(\oint \vec{E} \cdot d \vec{l} > 0\)

    • 在有限区域内有端点
    • 延伸到 \(\infty\)
    • 电场线密度

    很小的区域内,电场线认为是平行的。垂直取一个截面 \(d S\),

    \[ \text{“密度”} = \lim_{d S \to 0} \frac{\Phi}{|d S|} = \lim_{d S \to 0} \frac{\vec{E} \cdot d \vec{S}}{|d S|} = |\vec{E}| \]

微分形式

\[ \begin{aligned} \vec{E} \cdot d\vec{l} &= E_1 d x_1 + E_2 d x_2 + E_3 d x_3 \\ \vec{E} \cdot d \vec{S} &= E_1 d x_2 \wedge d x_3 + E_2 d x_3 \wedge d x_1 + E_3 d x_1 \wedge d x_2 \end{aligned} \]

\(d x_i, \, d x_i \wedge d x_j\) 视作线性空间的一组基,\(E_i\) 是坐标分量,这些分量承载了电场的信息。

\[ \begin{aligned} & \text{0-形式} && f(\vec{x}) \\ & \text{1-形式} && \sum \# \, d x_i \\ & \text{2-形式} && \sum \# \, d x_i \wedge d x_j \\ & \text{3-形式} && \# \, d x_1 \wedge d x_2 \wedge d x_3 \end{aligned} \]

三维空间中只有这些形式了,再加一个 \(\wedge\) 必然会变为 0. 这些形式之间如何转换?

考虑一个全微分 \(\mathrm{d} \boxed{\ldots dx} = \sum_i \frac{\partial \boxed{\ldots dx}}{\partial x_i} \boxed{?} dx_i\),问号处应该是楔积。于是引出外微分(exterior derivative,全微分的推广):

\[ \mathrm{d} \left(\sum_{i, j, k, \ldots} C_{i j k \ldots}(\vec{x}) \, d x_i \wedge d x_j \wedge d x_k \wedge \ldots \right) = \sum_a \sum_{i, j, k, \ldots} \frac{\partial C_{i j k \ldots}}{\partial x_a} \textcolor{orange}{d x_a \wedge} (d x_i \wedge d x_j \wedge d x_k \wedge \ldots) \]
  • \(0 \to 1\): \(\text{0-形式}: f(\vec{x})\)

    \[ \mathrm{d} \underset{\text{0-形式}}{\underbrace{f(\vec{x})}} \longrightarrow \underset{\text{1-形式}}{\underbrace{\sum_i \frac{\partial f}{\partial x_i} d x_i}} \]
  • \(1 \to 2\): \(\text{1-形式}: \omega^{(1)} = \sum_i^3 V_i(\vec{x}) d x_i\)

    \[ \begin{aligned} \mathrm{d} \omega^{(1)} &= (\partial_2 V_1 \, d x_2 \wedge d x_1 + \partial_3 V_1 \, d x_3 \wedge d x_1) &&= (\partial_2 V_3 - \partial_3 V_2)\; d x_2 \wedge d x_3 \\ &+ (\partial_1 V_2 \, d x_1 \wedge d x_2 + \partial_3 V_2 \, d x_3 \wedge d x_2) &&+ (\partial_3 V_1 - \partial_1 V_3)\; d x_3 \wedge d x_1 \\ &+ (\partial_1 V_3 \, d x_1 \wedge d x_3 + \partial_2 V_3 \, d x_2 \wedge d x_3) &&+ (\partial_1 V_2 - \partial_2 V_1)\; d x_1 \wedge d x_2 \end{aligned} \]

    注意到 \(d \vec{S} = (d x_2 \wedge d x_3, d x_3 \wedge d x_1, d x_1 \wedge d x_2)\),各分量正好是旋度的形式。

  • \(2 \to 3\): \(\text{2-形式}: \omega^{(2)} = u_1 \, d x_2 \wedge d x_3 + u_2 \, d x_3 \wedge d x_1 + u_3 \, d x_1 \wedge d x_2\)

    \[ d \omega^{(2)} = (\partial_1 u_1 + \partial_2 u_2 + \partial_3 u_3)\; d x_1 \wedge d x_2 \wedge d x_3 \]