3

Can anyone help me with aligning this long equation to the left please? Many thanks in advance. 

\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}
egreg
  • 1,121,712
Walter
  • 31

2 Answers2

2

You can use split, maybe with a different alignment point from = and a nested aligned:

\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}

Note the \supsub and \subsup commands for the stacked subscript and superscripts.

Also, \big should either be \bigl or \bigr, depending if it precedes an opening or a closing delimiter.

enter image description here

egreg
  • 1,121,712
1

enter image description here

\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}
gernot
  • 49,614