3

need help I need to put these three equations in three lines covered by a big parenthesis on the right side with the tag (Tolman 1939)

$$KP=e^{-\lambda} \Big( \frac {\nu'}{r}+ \frac {1}{r^2}\Big)-\frac {1}{r^2}$$

$$ KP= e^{-\lambda} \Big(\frac {\nu ''}{2}-\frac {\lambda ' \nu '}{4} + \frac {(\nu')^2}{4} + \frac {\nu' -\lambda'}{2r}\Big)$$

 $$K \rho c^2= e^{-\lambda}\Big (\frac {\lambda '}{r}-\frac {1}{r^2}\Big) +\frac {1}{r^2}$$
  • Comment aside: you should not use $$ … $$, which is plain TeX and can result in wrong vertical spacing. Use \[ … \] instead. – Bernard Aug 5 '14 at 10:05
4

You can add a brace on just one side with a \left. on one side and an \right\} to add the brace on the right.

enter image description here

References:

Code:

\documentclass{article}
\usepackage{amsmath}
\begin{document}

\begin{equation}
\left.
\begin{aligned}
             KP &= e^{-\lambda} \Big( \frac {\nu'}{r}+ \frac {1}{r^2}\Big)-\frac {1}{r^2} \\
             KP &= e^{-\lambda} \Big(\frac {\nu ''}{2}-\frac {\lambda ' \nu '}{4} + \frac {(\nu')^2}{4} + \frac {\nu' -\lambda'}{2r}\Big) \\
     K \rho c^2 &= e^{-\lambda}\Big (\frac {\lambda '}{r}-\frac {1}{r^2}\Big) +\frac {1}{r^2}
\end{aligned}
\right\} \tag{Tolman 1939}
\end{equation}
\end{document}
2

A variant with a slighly simpler code, thanks to the empheq package (that loads mathtools, from the same author, and amsmath):

\documentclass{article}
\usepackage[overload]{empheq}

\begin{document}

\begin{align*}[right = \empheqrbrace]
             KP &= e^{-\lambda} \Big( \frac {\nu'}{r}+ \frac {1}{r^2}\Big)-\frac {1}{r^2}  \\
             KP &= e^{-\lambda} \Big(\frac {\nu ''}{2}-\frac {\lambda ' \nu '}{4} + \frac {(\nu')^2}{4} + \frac {\nu' -\lambda'}{2r}\Big)  \tag{Tolman 1939}\\
     K \rho c^2 &= e^{-\lambda}\Big (\frac {\lambda '}{r}-\frac {1}{r^2}\Big) +\frac {1}{r^2}
%
\end{align*}
\end{document} 

enter image description here

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