删除分割环境中方程之间的大空格

Question

我是 Latex 新手，正在努力消除方程式之间以及方程式和文本之间的大空格。我正在使用 \documentclass{extarticle}，我写的方程式很大，看起来像

\begin{equation}

\begin{split}

Long equations taking more than half page

\end{split}

\end{equation}

我认为 latex 将所有方程式视为一个整体，这就是为什么当剩下半页时，方程式会移到下一页。（我正在使用 split 命令，因为它可以使用“&”轻松对齐方程式）请告诉我如何摆脱这个问题。

实际代码是

 \section{Diffeomorphism}
$L_o = m(\dot{x}^2)^\frac{1}{2}$\\[10pt]
$\dot{x}^2 = \dot{x_\mu}\dot{x^\mu} = \dot{x_0}\dot{x^0}+\dot{x_1}\dot{x^1}+.....................+\dot{x_{D-1}}\dot{x^{D-1}}$\\[15pt]
Action $S = \int_{-\infty}^{\infty} d\tau \; m(\dot{x}^2)^\frac{1}{2}$\\[10pt]
Diffeomorphism\\
$\tau \rightarrow \tau' = f(\tau) \simeq \tau -\epsilon(\tau)$\\
$f(\tau) = finite\; at\; \tau =0$\\
$f(\tau) \rightarrow 0$ as $\tau \rightarrow \pm \infty$\\
Action remains invariant under Diffeomorphism Transformations. We can prove it as follows:\\
\begin{equation}
\begin{split}
S' & = \int_{-\infty}^{\infty} d\tau' \; m\left[\left(\frac{dx^\mu}{d\tau'}\right)\left(\frac{dx_\mu}{d\tau'}\right)\right]^\frac{1}{2}\\[10pt]
&=\int_{-\infty}^{\infty} \frac{d\tau'}{d\tau}d\tau \; m\left[\left(\frac{dx^\mu}{d\tau}\right)\left(\frac{d\tau}{d\tau'}\right)\left(\frac{dx_\mu}{d\tau}\right)\left(\frac{d\tau}{d\tau'}\right)\right]^\frac{1}{2}\\[10pt]
&=\int_{-\infty}^{\infty} d\tau \left(\frac{d\tau'}{d\tau}\right)\left(\frac{d\tau}{d\tau'}\right) \; m\left[\left(\frac{dx^\mu}{d\tau}\right)\left(\frac{dx_\mu}{d\tau}\right)\right]^\frac{1}{2}\\[10pt]
&= \int_{-\infty}^{\infty} d\tau \; m(\dot{x}^2)^\frac{1}{2}\\[10pt]
&=S
\end{split}
\end{equation} 

输出为

Answer 1

简短回答：不，您不能在拆分环境中拆分方程。您可以使用 align。尝试使用 allowdisplaybreaks 命令。

\begingroup
\allowdisplaybreaks
\begin{align}
    ....
\end{align}
\endgroup

我希望它能有所帮助

Answer 2

我不知道我是否给了你一个很好的答案，我按照你的要求做了，但我把所有内容都写在了一页上。在这里，我添加了我的 MWE 示例以及一些额外的包，比如parskip对齐文本，并且改进了一些命令。

\documentclass{extarticle}
\usepackage{mathtools,amssymb}
\usepackage{parskip}
\begin{document}
\section{Diffeomorphism}
$L_0 = m(\dot{x}^2)^\frac{1}{2}$.
\vskip2pt
\[\dot{x}^2 = \dot{x_\mu}\dot{x}^\mu = \dot{x_0}\dot{x}^0+\dot{x_1}\dot{x}^1+\cdots\cdots\cdots+\dot{x}_{D-1}\dot{x}^{D-1}.\]
\vskip2pt
\textbf{Action} $S = \displaystyle \int_{-\infty}^{\infty} d\tau \; m(\dot{x}^2)^\frac{1}{2}$.
\vskip2pt
\textbf{Diffeomorphism}
$\tau \rightarrow \tau' = f(\tau) \simeq \tau -\epsilon(\tau)$.

$f(\tau) =\text{ finite at }\tau =0$.

$f(\tau) \rightarrow 0$ as $\tau \rightarrow \pm \infty$.

Action remains invariant under \textsl{Diffeomorphism Transformations}. We can prove it as follows:
\vspace{-.2cm}
\begin{equation}
\begin{split}
S' & = \int_{-\infty}^{\infty} d\tau' \; m\left[\left(\frac{dx^\mu}{d\tau'}\right)\left(\frac{dx_\mu}{d\tau'}\right)\right]^\frac{1}{2}\\
&=\int_{-\infty}^{\infty} \left(\frac{d\tau'}{d\tau}\right) d\tau \; m\left[\left(\frac{dx^\mu}{d\tau}\right)\left(\frac{d\tau}{d\tau'}\right)\left(\frac{dx_\mu}{d\tau}\right)\left(\frac{d\tau}{d\tau'}\right)\right]^\frac{1}{2}\\
&=\int_{-\infty}^{\infty} d\tau \left(\frac{d\tau'}{d\tau}\right)\left(\frac{d\tau}{d\tau'}\right) \; m\left[\left(\frac{dx^\mu}{d\tau}\right)\left(\frac{dx_\mu}{d\tau}\right)\right]^\frac{1}{2}\\
&= \int_{-\infty}^{\infty} d\tau \; m(\dot{x}^2)^\frac{1}{2}\\
&=S
\end{split}
\end{equation} 
\end{document}