1

The equation that I want to split is

$$\frac{\partial (r^2 A_{t, r})}{\partial r}+(\alpha^2 r^2-
\frac{4M}{\alpha r})^{-1} \frac{\partial A_{t, \phi}}{\partial \phi}=
-2q\delta(r-\frac{(4M)^{\frac{1}{3}} a}{\alpha})\delta(\cos\phi - 1)$$
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    Please make your code compilable (if possible), or at least complete it with \documentclass{...}, the required \usepackage's, \begin{document}, and \end{document}. That may seem tedious to you, but think of the extra work it represents for TeX.SX users willing to give you a hand. Help them help you: remove that one hurdle between you and a solution to your problem. – Skillmon May 02 '18 at 18:19
  • You could do something like \begin{align*} \frac{\partial (r^2 A_{t, r})}{\partial r}+(\alpha^2 r^2- \frac{4M}{\alpha r})^{-1} \frac{\partial A_{t, \phi}}{\partial \phi}=\\ -2q\delta(r-\frac{(4M)^{\frac{1}{3}} a}{\alpha})\delta(\cos\phi - 1) \end{align*} using the amsmath package. – Steven B. Segletes May 02 '18 at 18:20

2 Answers2

2

You could load the amsmath package and employ a multline* environment.

enter image description here

Note that I've increased the sizes of two pairs of round pairs viw \biggl( and \biggr).

Do also note that one shouldn't use $$ to initiate and terminate display-math mode in a LaTeX document. For more on this topic, please see the posting Why is \[ ... \] preferable to $$ ... $$?

\documentclass[twocolumn]{article} % use a narrow column width
\usepackage{amsmath} % for 'multline*' environment
\begin{document}
\begin{multline*}
\frac{\partial (r^2 A_{t, r})}{\partial r}
+\biggl(\alpha^2 r^2-\frac{4M}{\alpha r}\biggr)^{\!-1}\, 
\frac{\partial A_{t, \phi}}{\partial \phi}\\
=-2q\delta\biggl(r-\frac{(4M)^{1/3} a}{\alpha}\biggr)
\delta(\cos\phi - 1)
\end{multline*}
\end{document}
Mico
  • 506,678
2

with multlined from the package mathtools and shortness \pdv{...}{...} for partial derivative from the package physics:

\documentclass{article}
\usepackage{mathtools}
\usepackage{physics}

\begin{document}
\[
\begin{multlined}[0.6\linewidth]
\pdv{(r^2 A_{t, r})}{r} + 
    \left(\alpha^2 r^2 - \frac{4M}{\alpha r}\right)^{-1} 
    \pdv{A_{t, \phi}}{\phi}   \\
= -2q\delta\left(r-\frac{(4M)^{1/3} a}{\alpha}\right)\delta(\cos\phi - 1)
\end{multlined}
\]
\end{document}

enter image description here

Zarko
  • 296,517