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我正在尝试写一个等式，其中右侧是图像，左侧是数学表达式。唯一的问题是表达式太长，必须用多行分隔。我希望最终结果是数学表达式相对于图像居中。我得到的最接近的输出是以下代码：

\begin{equation}
\vcenter{\hbox{\includegraphics[scale=0.2]{Image}}} =

\begin{split}

\frac{i k_g}{\Lambda}\;\delta^{a_1a_2}\bigg[\eta^{\nu\sigma}\Big(k_1^\mu        k_2^\rho+k_1^\mu k_1^\rho\Big)+\eta^{\mu\sigma}\Big(k_1^\nu k_2^\rho+k_1^\nu   k_1^\rho\Big)\\
&\eta^{\nu\rho}\Big(k_1^\sigma k_2^\mu+k_2^\sigma k_2^\mu\Big)+\eta^{\mu\rho}\Big(k_1^\sigma k_2^\nu+k_2^\sigma k_2^\nu\Big)\\
-\eta^{\rho\sigma}\Big(k_1^\mu k_2^\nu+k_1^\nu k_2^\mu\Big)-k_1\cdot k_2\Big(\eta^{\mu\rho}\eta^{\nu\sigma}+\eta^{\mu\sigma}\eta^{\nu\rho}\Big)\bigg].

\end{split}
\end{equation}

但这是输出我得到的输出

如您所见，表达式与图像没有真正对齐，并且第二行部分超出了页面。我该如何正确解决这个问题？

答案1

使用aligned；可以使图片垂直居中gathered，比 更简洁\vcenter。

\documentclass{article}

\usepackage{amsmath}
\usepackage{graphicx}

\begin{document}

Text before.
\begin{equation}
\begin{gathered}
\includegraphics[width=4cm]{example-image}
\end{gathered}
=
\begin{aligned}[t]
&\frac{i k_g}{\Lambda}\delta^{a_1a_2}\bigl[
  \eta^{\nu\sigma}(k_1^\mu k_2^\rho+k_1^\mu k_1^\rho)\\
&+\eta^{\mu\sigma}(k_1^\nu k_2^\rho+k_1^\nu k_1^\rho)
  \eta^{\nu\rho}(k_1^\sigma k_2^\mu+k_2^\sigma k_2^\mu)\\
&+\eta^{\mu\rho}(k_1^\sigma k_2^\nu+k_2^\sigma k_2^\nu)
 -\eta^{\rho\sigma}(k_1^\mu k_2^\nu+k_1^\nu k_2^\mu)\\
&-k_1\cdot k_2(\eta^{\mu\rho}\eta^{\nu\sigma}+
  \eta^{\mu\sigma}\eta^{\nu\rho})\bigr].
\end{aligned}
\end{equation}
Text after.

\end{document}

在此处输入图片描述

adjustbox如果您希望有不同的相对位置，则另一种可能性是使用。

\documentclass{article}

\usepackage{amsmath}
\usepackage[export]{adjustbox}
\usepackage{graphicx}

\begin{document}

Text before.
\begin{equation}
\begin{gathered}
\setlength{\adjboxvtop}{1.2\baselineskip}
\includegraphics[width=4cm,valign=t]{example-image}
=
\begin{aligned}[t]
&\frac{i k_g}{\Lambda}\delta^{a_1a_2}\bigl[
  \eta^{\nu\sigma}(k_1^\mu k_2^\rho+k_1^\mu k_1^\rho)\\
&+\eta^{\mu\sigma}(k_1^\nu k_2^\rho+k_1^\nu k_1^\rho)
  \eta^{\nu\rho}(k_1^\sigma k_2^\mu+k_2^\sigma k_2^\mu)\\
&+\eta^{\mu\rho}(k_1^\sigma k_2^\nu+k_2^\sigma k_2^\nu)
 -\eta^{\rho\sigma}(k_1^\mu k_2^\nu+k_1^\nu k_2^\mu)\\
&-k_1\cdot k_2(\eta^{\mu\rho}\eta^{\nu\sigma}+
  \eta^{\mu\sigma}\eta^{\nu\rho})\bigr].
\end{aligned}
\end{gathered}
\end{equation}
Text after.

\end{document}

在此处输入图片描述

答案2

它能完成split. 编辑：哎呀！我忘了\smash[b]拍照了……

% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly 
                                 % declare the paper format.

\usepackage[T1]{fontenc}         % Not always necessary, but recommended.
% End of standard header.  What follows pertains to the problem at hand.

\usepackage{amsmath}
\usepackage{mwe}



\begin{document}

Text before.
\begin{equation}
    \begin{split}
        \smash[b]{\vcenter{\hbox{\includegraphics[scale=0.2]{image}}}}
                    = {}&
            \frac{i k_g}
                {\Lambda}\;\delta^{a_1a_2}\biggl[\eta^{\nu\sigma}\Bigl(k_1^\mu        
                k_2^\rho+k_1^\mu k_1^\rho\Bigr)\\
            &+\eta^{\mu\sigma}\Bigl(k_1^\nu k_2^\rho+k_1^\nu k_1^\rho\Bigr)
                \cdot \eta^{\nu\rho}\Bigl(k_1^\sigma k_2^\mu
            +k_2^\sigma k_2^\mu\Bigr)\\
            &+\eta^{\mu\rho}
                \Bigl(k_1^\sigma k_2^\nu+k_2^\sigma k_2^\nu\Bigr)\\
            &-\eta^{\rho\sigma}\Bigl(k_1^\mu k_2^\nu+k_1^\nu k_2^\mu\Bigr)-
                k_1\cdot k_2\Bigl(\eta^{\mu\rho}\eta^{\nu\sigma}+
                \eta^{\mu\sigma}\eta^{\nu\rho}\Bigr)\biggr].
    \end{split}
\end{equation}
Text after.

\end{document}

我选择了不同的点来打破方程式，在我看来，它们看起来好多了。还请注意，我已根据需要用 或 替换\Big，\Bigl对\Bigr也一样\bigg。最后，请注意如何指定对齐点。

这是输出：

代码输出

添加

OP 澄清说，想要的是不同的输出，即

期望输出

以我的拙见，这看起来比上面的代码产生的输出更差；无论如何，它也没有造成任何特殊问题：

% My standard header for TeX.SX answers:
\documentclass[a4paper]{article} % To avoid confusion, let us explicitly 
                                 % declare the paper format.

\usepackage[T1]{fontenc}         % Not always necessary, but recommended.
% End of standard header.  What follows pertains to the problem at hand.

\usepackage{amsmath}
\usepackage{mwe}



\begin{document}

Text before.
\begin{equation}
    \vcenter{\hbox{\includegraphics[scale=0.2]{image}}} =
        \begin{aligned}
            &\frac{i    k_g}
                {\Lambda}\;\delta^{a_1a_2}\biggl[\eta^{\nu\sigma}\Bigl(k_1^\mu        
                k_2^\rho+k_1^\mu k_1^\rho\Bigr)\\
            &+\eta^{\mu\sigma}\Bigl(k_1^\nu k_2^\rho+k_1^\nu k_1^\rho\Bigr)
                \cdot \eta^{\nu\rho}\Bigl(k_1^\sigma k_2^\mu
            +k_2^\sigma k_2^\mu\Bigr)\\
            &+\eta^{\mu\rho}
                \Bigl(k_1^\sigma k_2^\nu+k_2^\sigma k_2^\nu\Bigr)\\
            &-\eta^{\rho\sigma}\Bigl(k_1^\mu k_2^\nu+k_1^\nu k_2^\mu\Bigr)-
                k_1\cdot k_2\Bigl(\eta^{\mu\rho}\eta^{\nu\sigma}+
                \eta^{\mu\sigma}\eta^{\nu\rho}\Bigr)\biggr].
        \end{aligned}
\end{equation}
Text after.

\end{document}

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