将长方程拆分为两行

Question

我正在尝试写一个长方程，但无法在不破坏对齐的情况下将其分成两个对齐

\documentclass[12pt,letterpaper]{article}
\usepackage[utf8]{inputenc}
\usepackage{titling}
\usepackage[a4paper,left=25mm,top=25mm,right=25mm,bottom=25mm]{geometry}
\title{\vspace{-2.5cm}}
\author{\vspace{-2.0cm}}
\date{\vspace{-2.0cm}}
\begin{document}
\[ds^2=-d\tau^2\bigg[1+\sum_{n=1}^{m}\frac{\sigma^{(n)}(\phi)} 
{\tau^n}\bigg]^2+2\tau d\tau 
d\phi^a\bigg[\sum_{n=1}^{m}\frac{A^{(n)}_a(\phi)}{\tau^n}\bigg]+\tau^2 
d\phi^a d\phi^b \bigg[h^{(0)}_{ab}+\sum_{n=1}^{m}\frac{h^{(n)}_{ab}} 
{\tau^n}\bigg]\]
Coordinate transformation

\[\phi^a = \bar{\phi}^a +\frac{1}{\bar{\tau}} G^{(1)a}\]
\[\tau=\bar{\tau}\]
Then we get,
\[d\phi^a=d\bar{\phi}^a+\frac{1}{\bar{\tau}}{\partial 
_b}G^{(1)a}d\bar{\phi}^b-\frac{1}{\bar{\tau^2}}d\bar{\tau}G^{(1)a}\]
Substituting this into the metric, we obtain

\[ds^2 = -d\bar{\tau}^2\bigg[1+\sum_{n=1}^{m}\frac{\sigma^{(n)}(\phi)} 
{\bar{\tau}^n}\bigg]^2+2d\bar{\tau}d\bar{\tau}\bigg[d\phi^a + \frac{1} 
{\bar{\tau}}\partial_c G^a d\phi^c - \frac{1} 
{\bar{\tau}^2}d\bar{\tau}G^a\bigg] \bigg[\sum_{n=1}^{m}\frac{A^{(n)}_a} 
{\bar{\tau}^n}\bigg] + \bar{\tau}^2\bigg[h^{(0)}_{ab} + 
\sum_{n=1}^{m}\frac{h^{(n)}_{ab}}{\bar{\tau}^n}\bigg] 
\bigg[d\bar{\phi}^a+\frac{1}{\bar{\tau}}\partial_cG^ad\bar{\phi}^c-\frac{1} 
{\bar{\tau}^2}d\bar{\tau}G^a\bigg]\bigg[d\bar{\phi}^b+\frac{1} 
{\bar{\tau}}\partial_jG^bd\bar{\phi}^j-\frac{1} 
{\bar{\tau}^2}d\bar{\tau}G^b\bigg]\]

\end{document}

Answer 1

使用 rheamsmath包：

\documentclass[12pt,letterpaper]{article}
\usepackage[a4paper,left=25mm,top=25mm,right=25mm,bottom=25mm]{geometry}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}

\begin{document}
    \[
ds^2=-d\tau^2\biggl[1+\sum_{n=1}^{m}\frac{\sigma^{(n)}(\phi)}
{\tau^n}\biggr]^2+2\tau d\tau
d\phi^a\biggl[\sum_{n=1}^{m}\frac{A^{(n)}_a(\phi)}{\tau^n}\biggr]+\tau^2
d\phi^a d\phi^b \biggl[h^{(0)}_{ab}+\sum_{n=1}^{m}\frac{h^{(n)}_{ab}}
{\tau^n}\biggr]
    \]
Coordinate transformation
    \begin{align*} % <---
\phi^a & = \bar{\phi}^a +\frac{1}{\bar{\tau}} G^{(1)a}\\
  \tau & = \bar{\tau}
    \end{align*}
Then we get,
   \[
d\phi^a=d\bar{\phi}^a+\frac{1}{\bar{\tau}}{\partial
_b}G^{(1)a}d\bar{\phi}^b-\frac{1}{\bar{\tau^2}}d\bar{\tau}G^{(1)a}
    \]
Substituting this into the metric, we obtain
    \begin{multline*} % <---
ds^2 = -d\bar{\tau}^2\biggl[1+\sum_{n=1}^{m}\frac{\sigma^{(n)}(\phi)}
    {\bar{\tau}^n}\biggr]^2+2d\bar{\tau}d\bar{\tau}\biggl[d\phi^a + \frac{1}
    {\bar{\tau}}\partial_c G^a d\phi^c
    - \frac{1}{\bar{\tau}^2}d\bar{\tau}G^a\biggr] \biggl[\sum_{n=1}^{m}\frac{A^{(n)}_a}
    {\bar{\tau}^n}\biggr] \\
+ \bar{\tau}^2\biggl[h^{(0)}_{ab} + \sum_{n=1}^{m}\frac{h^{(n)}_{ab}}{\bar{\tau}^n}\biggr]
\biggl[d\bar{\phi}^a+\frac{1}{\bar{\tau}}\partial_cG^ad\bar{\phi}^c-\frac{1}
{\bar{\tau}^2}d\bar{\tau}G^a\biggr]\biggl[d\bar{\phi}^b+\frac{1}
{\bar{\tau}}\partial_jG^bd\bar{\phi}^j-\frac{1}
{\bar{\tau}^2}d\bar{\tau}G^b\biggr]
    \end{multline*}

\end{document}

如果d表示微分（而不是变量），则应考虑将其写成直立形式\mathrm{d}.

Answer 2

这是一个使用单一align*环境的单独解决方案。这允许对=符号进行对齐。

\documentclass[12pt,letterpaper]{article} % or "a4paper", as appropriate
\usepackage[margin=25mm]{geometry}
\usepackage{amsmath} % for "align*" env.
\begin{document}
\allowdisplaybreaks

\begin{align*}
ds^2 &= -d\tau^2
\biggl[1+\sum_{n=1}^{m}\frac{\sigma^{(n)}(\phi)}
{\tau^n}\biggr]^2
+2\tau d\tau d\phi^a
\biggl[\sum_{n=1}^{m}\frac{A^{(n)}_a(\phi)}{\tau^n}\biggr]
+\tau^2 d\phi^a d\phi^b \biggl[h^{(0)}_{ab}+\sum_{n=1}^{m}\frac{h^{(n)}_{ab}}
{\tau^n}\biggr] \\
\intertext{Coordinate transformation:}
\phi^a &= \bar{\phi}^a +\frac{1}{\bar{\tau}} G^{(1)a}\\
\tau   &=\bar{\tau}\\
\intertext{We then get}
d\phi^a &= d\bar{\phi}^a+\frac{1}{\bar{\tau}}{\partial
_b}G^{(1)a}d\bar{\phi}^b-\frac{1}{\bar{\tau}^2}d\bar{\tau}G^{(1)a} \\
\intertext{Substituting this into the metric, we obtain}
ds^2 &= -d\bar{\tau}^2
\biggl[1+\sum_{n=1}^{m}\frac{\sigma^{(n)}(\phi)}
{\bar{\tau}^n}\biggr]^2
+2d\bar{\tau}d\bar{\tau}
\biggl[d\phi^a + \frac{1}{\bar{\tau}} \partial_c G^a d\phi^c 
- \frac{1}{\bar{\tau}^2}d\bar{\tau}G^a\biggr]
\biggl[\sum_{n=1}^{m}\frac{A^{(n)}_a}
{\bar{\tau}^n}\biggr] \\
&\qquad + \bar{\tau}^2\biggl[h^{(0)}_{ab} +
\sum_{n=1}^{m}\frac{h^{(n)}_{ab}}{\bar{\tau}^n}\biggr]
\biggl[d\bar{\phi}^a+\frac{1}{\bar{\tau}}\partial_cG^ad\bar{\phi}^c
-\frac{1}{\bar{\tau}^2}d\bar{\tau}G^a\biggr]
\biggl[d\bar{\phi}^b+\frac{1}{\bar{\tau}} 
\partial_jG^bd\bar{\phi}^j
-\frac{1}{\bar{\tau}^2}d\bar{\tau}G^b\biggr]
\end{align*}

\end{document}