证明出现在方程的中间

Question

有人知道为什么会发生这种情况吗？

\begin{proof}
\begin{aligned} 
\bar{p} \cdot \bar{x} 
& =\bar{p}^\mu \bar{x}_\mu \\ 
& =\Lambda^\mu{ }_\nu p^\nu \Lambda_\mu{ }^\rho x_p \\ 
& =\left(\Lambda^{-1}\right)^\rho{ }_\mu \Lambda^\mu{ }_\nu p^\nu x_p \\ 
& =\delta_\nu^\rho p^\nu x_\rho \\
& =p^\nu x_\nu \\ & =p \cdot x

\end{aligned}
\end{proof}

Answer 1

aligned必须处于数学模式，并且您想要顶部对齐，但我会align*在这里使用。

\documentclass{article}

\usepackage{amsmath,amsthm}

\begin{document}

aligned c
\begin{proof}
$\begin{aligned} 
\bar{p} \cdot \bar{x} 
& =\bar{p}^\mu \bar{x}_\mu \\ 
& =\Lambda^\mu{ }_\nu p^\nu \Lambda_\mu{ }^\rho x_p \\ 
& =\left(\Lambda^{-1}\right)^\rho{ }_\mu \Lambda^\mu{ }_\nu p^\nu x_p \\ 
& =\delta_\nu^\rho p^\nu x_\rho \\
& =p^\nu x_\nu \\ & =p \cdot x
\end{aligned}$
\end{proof}

aligned t
\begin{proof}
$\begin{aligned}[t] 
\bar{p} \cdot \bar{x} 
& =\bar{p}^\mu \bar{x}_\mu \\ 
& =\Lambda^\mu{ }_\nu p^\nu \Lambda_\mu{ }^\rho x_p \\ 
& =\left(\Lambda^{-1}\right)^\rho{ }_\mu \Lambda^\mu{ }_\nu p^\nu x_p \\ 
& =\delta_\nu^\rho p^\nu x_\rho \\
& =p^\nu x_\nu \\ & =p \cdot x
\end{aligned}$
\end{proof}


Align
\begin{proof}
\begin{align*}
\bar{p} \cdot \bar{x} 
& =\bar{p}^\mu \bar{x}_\mu \\ 
& =\Lambda^\mu{ }_\nu p^\nu \Lambda_\mu{ }^\rho x_p \\ 
& =\left(\Lambda^{-1}\right)^\rho{ }_\mu \Lambda^\mu{ }_\nu p^\nu x_p \\ 
& =\delta_\nu^\rho p^\nu x_\rho \\
& =p^\nu x_\nu \\ & =p \cdot x \qedhere
\end{align*}
\end{proof}

\end{document}