答案1
我会用tikzmark
。
\documentclass{article}
\usepackage{tikz,amsmath}
\usetikzlibrary{tikzmark}
\usepackage{lipsum} %>>> for dummy text only
\begin{document}
\lipsum[1]
\vspace*{8mm} % a gap for tikzmark later. Change as you wish
\[A(r)=\int_{V'}\dfrac{\mu_0}{4\pi}\dfrac{M(r')\times \hat{R} d\tau'}{R^2}\;
\tikzmarknode{e1}{=}
\;\dfrac{\mu_0}{4\pi}\int_{V'}M(r')\times\nabla'\left(\dfrac{1}{R}\right)d\tau'
\]
\begin{tikzpicture}[overlay,remember picture]
\draw[stealth-,cyan,thick] (e1)+(90:.2)--+(90:.7) node[above=-2mm]{$\dfrac{\hat R}{R^2}=\nabla'\left(\dfrac{1}{R}\right)$};
\end{tikzpicture}
\lipsum[2]
\end{document}
答案2
我不会在已经很长的推导中塞入太多视觉信息,而是使用\shortintertext{...}
和\intertext{...}
指令将解释与推导的实际步骤穿插在一起。这样,你就可以让读者在步骤之间有一些视觉上的“喘息”。如果你觉得有必要或有益,你仍然可以使用颜色(例如蓝色)将解释与实际公式在视觉上区分开来。
\documentclass{article}
\usepackage{mathtools,bm,xcolor}
\begin{document}
\begin{align*}
\bm{A}(\bm{r})
&= \int\limits_{V'} \frac{\mu_0}{4\pi} \frac{\bm{M}(\bm{r}') \times \widehat{\bm{R}}\, d\tau'}{R^2}\\
\shortintertext{Employ \textcolor{blue}{$\displaystyle\frac{\widehat{\bm{R}}}{R^2}=\bm{\nabla}'\biggl(\frac{1}{R}\biggr)$}:}
&= \frac{\mu_0}{4\pi} \int\limits_{V'} \bm{M}(\bm{r}')\times \bm{\nabla}'\biggl(\frac{1}{R}\biggr) \, d\tau'\\
\intertext{Employ \textcolor{blue}{$\displaystyle \bm{M}(\bm{r}')\times \bm{\nabla}'\biggl(\frac{1}{R}\biggr) =
\frac{\bm{\nabla}'\times \bm{M}}{R}-\bm{\nabla}'\times\biggl(\frac{\bm{M}}{R}\biggr)$}:}
&= \frac{\mu_0}{4\pi}\int\limits_{V'} \frac{\bm{\nabla}'\times \bm{M}}{R} \,d\tau' +
\frac{\mu_0}{4\pi}\int\limits_{V'}\biggl[-\bm{\nabla}'\times\biggl(\frac{\bm{R}}{R} \biggr) \biggr]\,d\tau'\\
\intertext{Employ \dots}
\end{align*}
\end{document}