Group displayed equations with braces
You can use eqparbox
. The label to use should be unique to the group of boxes to typeset with the same width.
I'd prefer left alignment, change to \eqmath[c]{A}{...}
if you want centering.
\documentclass{article}
\usepackage{amsmath}
\usepackage{empheq}
\usepackage{eqparbox}
\DeclareMathOperator{\Div}{div}
\DeclareMathOperator{\Rot}{rot}
\newcommand{\eqmath}[3][l]{%
% #1 = alignment, default l, #2 = label, #2 = math material
\eqmakebox[#2][#1]{$\displaystyle#3$}%
}
\begin{document}
Maxwell's equations:
\begin{empheq}[left=\empheqlbrace]{align}
&\eqmath{A}{\Div B = 0\vphantom{\Big|}} \tag{Thomson}\\
&\eqmath{A}{\Rot E + \frac{\partial B}{\partial t} = 0} \tag{Faraday}
\end{empheq}
\begin{empheq}[left=\empheqlbrace]{align}
&\eqmath{A}{\Div E = \frac{\rho}{\epsilon_0}} \tag{Gauss}\\
&\eqmath{A}{\Rot B - \frac{1}{c^2} \frac{\partial E}{\partial t} = \mu_0 J} \tag{Amp\`ere}
\end{empheq}
\end{document}
Please, note that it should be
\frac{1}{c^2}
True, \frac12
works, but…
Actually, I'd prefer not to abuse \tag
.
\documentclass{article}
\usepackage{amsmath}
\usepackage{eqparbox}
\DeclareMathOperator{\Div}{div}
\DeclareMathOperator{\Rot}{rot}
\newcommand{\eqmath}[3][l]{%
% #1 = alignment, default l, #2 = label, #2 = math material
\eqmakebox[#2][#1]{$\displaystyle#3$}%
}
\newcommand{\eqtext}[3][l]{%
% #1 = alignment, default l, #2 = label, #2 = text material
\eqmakebox[#2][#1]{#3}%
}
\begin{document}
Maxwell's equations:
\begin{gather*}
\left\{\begin{alignedat}{2}
&\eqmath{B}{\Div B = 0\vphantom{\Big|}} &\quad&\eqtext{C}{(Thomson)}\\
&\eqmath{B}{\Rot E + \frac{\partial B}{\partial t} = 0} &&\eqtext{C}{(Faraday)}
\end{alignedat}\right.
\\[1ex]
\left\{\begin{alignedat}{2}
&\eqmath{B}{\Div E = \frac{\rho}{\epsilon_0}} &\quad&\eqtext{C}{(Gauss)}\\
&\eqmath{B}{\Rot B - \frac{1}{c^2} \frac{\partial E}{\partial t} = \mu_0 J} &&\eqtext{C}{(Amp\`ere)}
\end{alignedat}\right.
\end{gather*}
\end{document}
With labels before the braces, both realizations:
\documentclass{article}
\usepackage{amsmath}
\usepackage{empheq}
\usepackage{eqparbox}
\DeclareMathOperator{\Div}{div}
\DeclareMathOperator{\Rot}{rot}
\newcommand{\eqmath}[3][l]{%
% #1 = alignment, default l, #2 = label, #2 = math material
\eqmakebox[#2][#1]{$\displaystyle#3$}%
}
\newcommand{\eqtext}[3][l]{%
% #1 = alignment, default l, #2 = label, #2 = text material
\eqmakebox[#2][#1]{#3}%
}
\begin{document}
\section{With empheq}
Maxwell's equations:
\begin{empheq}[left={\makebox[4em][r]{Source-free }\empheqlbrace}]{align}
&\eqmath{A}{\Div B = 0\vphantom{\Big|}} \tag{Thomson}\\
&\eqmath{A}{\Rot E + \frac{\partial B}{\partial t} = 0} \tag{Faraday}
\end{empheq}
\begin{empheq}[left={\makebox[4em][r]{With sources }\empheqlbrace}]{align}
&\eqmath{A}{\Div E = \frac{\rho}{\epsilon_0}} \tag{Gauss}\\
&\eqmath{A}{\Rot B - \frac{1}{c^2} \frac{\partial E}{\partial t} = \mu_0 J} \tag{Amp\`ere}
\end{empheq}
\section{No empheq}
Maxwell's equations:
\begin{gather*}
\eqtext[r]{1}{Source-free }
\left\{\begin{alignedat}{2}
&\eqmath{B}{\Div B = 0\vphantom{\Big|}} &\quad&\eqtext{C}{(Thomson)}\\
&\eqmath{B}{\Rot E + \frac{\partial B}{\partial t} = 0} &&\eqtext{C}{(Faraday)}
\end{alignedat}\right.
\\[1ex]
\eqtext[r]{1}{With sources }
\left\{\begin{alignedat}{2}
&\eqmath{B}{\Div E = \frac{\rho}{\epsilon_0}} &\quad&\eqtext{C}{(Gauss)}\\
&\eqmath{B}{\Rot B - \frac{1}{c^2} \frac{\partial E}{\partial t} = \mu_0 J} &&\eqtext{C}{(Amp\`ere)}
\end{alignedat}\right.
\end{gather*}
\end{document}
If you don't need to cross-reference the equations, you have a solution with the bigdelim
package and tabularx
. I added the esdiff
package for an easy typing of partial derivatives, and cellspace
to add some vertical padding to the table rows:
\documentclass{article}
\usepackage{esdiff, empheq}
\DeclareMathOperator{\Div}{div}
\DeclareMathOperator{\rot}{rot}
\usepackage{tabularx, bigdelim}
\usepackage[column=O]{cellspace}
\setlength{\cellspacetoplimit}{4pt}
\setlength{\cellspacebottomlimit}{4pt}
\begin{document}
Maxwell's equations:
\noindent\begin{tabularx}{\textwidth}{@{}Xr @{}>{$\displaystyle}Ol<{$}>{\raggedleft\arraybackslash}X @{}}
&\ldelim\{{3}{*}[Source-free~] & \Div B = 0 & \text{(Thomson)}\\
& & \rot E + \diffp{B}{t} = 0 &\text{(Faraday)} \\ \strut \\
&\ldelim\{{4}{*}[With sources~] & \Div E = \frac{\rho}{\epsilon} & \text{(Gauss)}\\
& & \rot B - \frac1{c^2} \diffp{E}{t} = \mu_0 J &\text{(Ampère)}
\end{tabularx}
\end{document}