How can I construct three vertical dots similar to \vdots but which span the same height as a colon :? \vdots is usually intended for use in matrices or a set of equations to indicate that there are many elements. I am intending its use within an in-line equation for purposes of operator ordering.
[Tex/LaTex] Vertical dots with same height as colon
math-modesymbols
Related Solutions
An alignat* environment is more suitable here than align* or an array (within a display-math environment). See below.
For numbering lines, you can use the command \numberthis which I define below. That command takes one optional argument, which if not empty, is used to create a label for cross-referencing. (Thanks to wasteofspace for his/her comments.)
Also, see David Carlisle's recommendation regarding the (ab)use of \ensuremath.
\documentclass[11pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[landscape]{geometry}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{mathtools}
\newcommand\numberthis[1][]{%
\refstepcounter{equation}%
\ifx#1\empty\else\label{eq:#1}\fi%
\tag{\theequation}%
}
\newcommand*{\defeq}{\vcentcolon=}
\newcommand*{\diff}{\mathrm{d}}
\newcommand*{\pd}[3][]{%
\frac{\partial^{#1}#2}{\partial #3^{#1}}
}
\newcommand*{\pdc}[4][]{%
\left(\pd[#1]{#2}{#3}\right)_{\!\!#4}
}
\newcommand{\td}[3][]{%
\frac{\diff^{#1}#2}{\diff #3^{#1}}
}
\newcommand{\earray}[3][2]{%
\everymath{\displaystyle}%
\renewcommand*{\arraystretch}{#1}%
\begin{array}{#2}
#3
\end{array}
}
\begin{document}
\begin{alignat*}{4}
&\diff J^\pm
&& \defeq \diff u\pm \frac{1}{C}\diff p
&& =\left(F\pm\frac{1}{C}G\pdc{p}{e}{V}\right)\diff t
&& \text{ along a path defined by } \td{\xi}{t}=\pm C,\\[1em]
&\diff S
&& \defeq \diff p + C^2 \diff V
&& =G\pdc{p}{e}{V} \diff t
&& \text{ along a path defined by } \td{\xi}{t}=0 \numberthis[ju]
\end{alignat*}
Equation \ref{eq:ju} is of interest for\ldots
\end{document}
These are obviously binary operators, so they should carry the same spacing. That is, use whatever works and then wrap it in \mathbin. While the original picture showed the bottom dots resting on the baseline, I think it would be more correct to center the symbols on the math axis (where the \cdot is placed). Here is a simple possibility, that does not allow for size changes in scripts, but does respect global size changes like \large.
\documentclass{article}
\def\onedot{$\mathsurround0pt\ldotp$}
\def\cddot{% two dots stacked vertically
\mathbin{\vcenter{\baselineskip.67ex
\hbox{\onedot}\hbox{\onedot}}%
}}%
\def\cdddot#1{% three dots
\mathbin{\vcenter{\baselineskip.67ex
\hbox{\onedot}\hbox{\onedot}\hbox{\onedot}%
}}%
}
\begin{document}
Test it:
\[
\mathbf{P} = \varepsilon_0 \left(\chi^{(1)}
\cdot \mathbf{E} + \chi^{(2)}
\cddot \mathbf{EE} + \chi^{(3)}
\cdddot \mathbf{EEE} + \cdots \right)
\]
\end{document}
Allowing for changes in scripts is a good deal harder, but could be done with the aid of some of the internals of the mathdots package.
Best Answer
If you want a version of
\vdotsthat is exactly as tall as a colon you could try to overlay a\cdoton top of a:. The code below, which is based on this excellent answer, does exactly this.I'm actually defining several different versions of
\threedotswith different spacing properties:\threedotsordhas the same spacing as ordinary letters;\threedotsopenand\threedotsclosehave the same spacing as opening and closing parentheses respectively;\threedotsbinhas the same spacing as a binary operator (like+,-,\times,…);\threedotsrelhas the same spacing as a relation symbol (like=,<,\sim,\rightorrow, …).Which version you need will depend on how you intend to use the symbol. Since you're talking about operator ordering, you will probably want to use
\threedotsopenand\threedotsclose. For convenience, I defined a macro\oporderthat puts a pair of triple dots with the correct spacing around its argument.Note that
\mathopvertically centres its argument with respect to the math axis whenever it is applied to a single character, as remarked e.g. here. You could also centre the\cdotwith respect to the:, but this is more work (if you want to do it automatically, without guessing the amount to lower it by).