长公式需要拆分并左对齐

Question

有人能帮我把这个长方程式向左对齐吗？非常感谢。

\begin{equation} \label{334}
 \begin{split}
\theta_{\mu\nu}(x)=T_{\mu\nu}(x)
+\frac{1}{D-2}\big(\partial_{\mu}\partial_{\alpha}L^{\alpha}_{\;\;\nu}(x)
+\partial_{\nu}\partial_{\alpha}L^{\alpha}_{\;\;\mu}(x)-\\
\partial^{2}L_{\mu\nu}(x)-g_{\mu\nu}\partial_{\alpha}\partial_{\beta}L^{\alpha\beta}(x)\big)
+\frac{1}{(D-2)(D-1)}\big(g_{\mu\nu}\partial^{2}L_{\alpha}^{\;\;\alpha}(x)-\partial_{\mu}\partial_{\nu}L_{\alpha}^{\;\;\alpha}(x)\big),
\end{split}
\end{equation}

Answer 1

您可以使用split，也许使用来自的对齐点=和嵌套对齐：

\documentclass{article}
\usepackage{amsmath,array}

\newcommand{\supsub}[2]{%
  ^{#1}_{\hphantom{#1}#2}%
}
\newcommand{\subsup}[2]{%
  _{#1}^{\hphantom{#1}#2}%
}

\begin{document}

\begin{equation} \label{334}
\begin{split}
\theta_{\mu\nu}(x)=T_{\mu\nu}(x)
  &+\begin{aligned}[t]
    \frac{1}{D-2}\bigl(&\partial_{\mu}\partial_{\alpha}L\supsub{\alpha}{\nu}(x)
                       +\partial_{\nu}\partial_{\alpha}L\supsub{\alpha}{\mu}(x)\\
                       &-\partial^{2}L_{\mu\nu}(x)
                        -g_{\mu\nu}\partial_{\alpha}\partial_{\beta}L^{\alpha\beta}(x)\bigr)
    \end{aligned}
  \\
  &+\frac{1}{(D-2)(D-1)}\bigl(g_{\mu\nu}\partial^{2}L\subsup{\alpha}{\alpha}(x)
   -\partial_{\mu}\partial_{\nu}L\subsup{\alpha}{\alpha}(x)\bigr),
\end{split}
\end{equation}

\end{document}

注意堆叠下标和上标的\supsub和命令。\subsup

此外，\big应该是\bigl或\bigr，取决于它是位于开始分隔符还是结束分隔符之前。

Answer 2

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{alignat}{2}
\theta_{\mu\nu}(x)={}&T_{\mu\nu}(x)
+\frac{1}{D-2}\big(&&
  \partial_{\mu}\partial_{\alpha}L^{\alpha}_{\;\;\nu}(x)
   +\partial_{\nu}\partial_{\alpha}L^{\alpha}_{\;\;\mu}(x)\notag\\
  &&&-\partial^{2}L_{\mu\nu}(x)-g_{\mu\nu}\partial_{\alpha}\partial_{\beta}L^{\alpha\beta}(x)\big)\notag\\
&\lefteqn{{}+\frac{1}{(D-2)(D-1)}\big(g_{\mu\nu}\partial^{2}L_{\alpha}^{\;\;\alpha}(x)-\partial_{\mu}\partial_{\nu}L_{\alpha}^{\;\;\alpha}(x)\big),}
\label{334}
\end{alignat}
\end{document}