在等式中插入文本

Question

我尝试在公式中插入文本，但文档中没有正确介绍这种方法amsmath。这些解决方案包括在公式之间插入文本，而我尝试将其插入到对齐语句的中间：

\begin{align}
\mathfrak{E}_x^e+\mathfrak{E}_x^b = 0, \quad \mathfrak{E}_y^e&+\mathfrak{E}_y^b = 0, \quad \mathfrak{B}_z^e+\mathfrak{B}_z^b = 0
\shortintertext{on F,} 
\frac{\partial}{\partial z} (\mathfrak{E}_z^e+\mathfrak{E}_z^b) = 0, \quad \frac{\partial}{\partial z}  (\mathfrak{B}_x^e&+\mathfrak{B}_x^b) = 0,\quad \frac{\partial}{\partial z} (\mathfrak{E}_z^e+\mathfrak{E}_z^b) = 0
\end{align}

编译结果如下：

我希望这两个语句都被编号为 (5)，也就是说，数字 5 出现在中间的行上（在 F, 上），而另外两个数字则消失。

Answer 1

这是一个非常肮脏的伎俩，一旦你改变编号方案，它就会像玻璃一样破碎。

\documentclass{article}
\usepackage{mathtools}
\usepackage{showframe}
\def\mathfrak#1{#1}%<- I simply cannot remember which package is needed, and it is not relevant to the question.
\begin{document}
\begin{align}
    \mathfrak{E}_x^e+\mathfrak{E}_x^b = 0, \quad
    \mathfrak{E}_y^e&+\mathfrak{E}_y^b = 0, \quad
    \mathfrak{B}_z^e+\mathfrak{B}_z^b = 0
    \shortintertext{on F,}
    \frac{\partial}{\partial z}
    (\mathfrak{E}_z^e+\mathfrak{E}_z^b) = 0, \quad
    \frac{\partial}{\partial z}
    (\mathfrak{B}_x^e&+\mathfrak{B}_x^b) = 0,\quad
    \frac{\partial}{\partial z}
    (\mathfrak{E}_z^e+\mathfrak{E}_z^b) = 0
\end{align}
\begin{align*}
    \refstepcounter{equation}
    \mathfrak{E}_x^e+\mathfrak{E}_x^b = 0, \quad
    \mathfrak{E}_y^e&+\mathfrak{E}_y^b = 0, \quad
    \mathfrak{B}_z^e+\mathfrak{B}_z^b = 0
    \shortintertext{on F,\hfill(\theequation)}
    \frac{\partial}{\partial z}
    (\mathfrak{E}_z^e+\mathfrak{E}_z^b) = 0, \quad
    \frac{\partial}{\partial z}
    (\mathfrak{B}_x^e&+\mathfrak{B}_x^b) = 0,\quad
    \frac{\partial}{\partial z}
    (\mathfrak{E}_z^e+\mathfrak{E}_z^b) = 0
\end{align*}
\end{document}

Answer 2

这是另一个肮脏的伎俩。（不确定你为什么使用align；gather对我来说似乎更合适。但aligned在这里也应该有效。）

\documentclass{article}
\usepackage{mathtools,amsfonts}
\begin{document}
\begin{align}
\mathfrak{E}_x^e+\mathfrak{E}_x^b = 0, \quad \mathfrak{E}_y^e&+\mathfrak{E}_y^b = 0, \quad \mathfrak{B}_z^e+\mathfrak{B}_z^b = 0
\shortintertext{on F,} 
\frac{\partial}{\partial z} (\mathfrak{E}_z^e+\mathfrak{E}_z^b) = 0, \quad \frac{\partial}{\partial z}  (\mathfrak{B}_x^e&+\mathfrak{B}_x^b) = 0,\quad \frac{\partial}{\partial z} (\mathfrak{E}_z^e+\mathfrak{E}_z^b) = 0
\end{align}
some text
\begin{equation}
\begin{gathered}
\mathfrak{E}_x^e+\mathfrak{E}_x^b = 0, \quad \mathfrak{E}_y^e +\mathfrak{E}_y^b = 0, \quad \mathfrak{B}_z^e+\mathfrak{B}_z^b = 0\\
\makebox[.5\textwidth]{\llap{\kern-.5\textwidth\rlap{on F,}\hfil}\null\hfill}\\
\frac{\partial}{\partial z} (\mathfrak{E}_z^e+\mathfrak{E}_z^b) = 0, \quad \frac{\partial}{\partial z}  (\mathfrak{B}_x^e +\mathfrak{B}_x^b) = 0,\quad \frac{\partial}{\partial z} (\mathfrak{E}_z^e+\mathfrak{E}_z^b) = 0
\end{gathered}
\end{equation}
\end{document}