Slight misalignment using hphantom equals sign in align

Question

I am attempting to align the integral signs precisely, whilst still retaining the correct spacing around the equals signs. Why doesn't the following code get it exactly right?

\begin{align}
    & \hphantom{{}={}} \int \\
    & = \int \\
    & = F
\end{align}

enter image description here

Does this answer your question? Suppressing = on a line in the align environment — user202729, Jul 27 '22 at 05:48
I asked this question 7 years ago, but it does not appear to be the same as the one linked. — bdforbes, Jul 27 '22 at 11:50
Default flag-generated comment for possible duplicate flag. The one with close/reopen voting privilege are the one making the decision, not me. — user202729, Jul 27 '22 at 12:03

score 12 · Answer 1 · answered May 28 '15 at 06:27

Let's start with an analysis of math atoms and spacings between them:

& = \int

The align environment adds an invisible empty subformula {} at the start of the column:

{} = \int

A subformula is acts as \mathord. The following line shows the math atoms with their type in the subscript position:

{}_ord =_rel \int_op

There is a table in "The TeXbook" from Donald E. Knuth or in section "23.6.1 Classification of mu glue" in book "TeX by Topic" from Victor Eijkhout, which shows the spacing between math atoms. The spacing information is put in the superscript position between the math atoms:

{}_ord ^thick =_rel ^thick \int_op ^thin \int_op

The same analysis for

& \hphantom{{}={}} \int

yields:

{}_ord ^none { {}_ord ^thick =_rel ^thick {}_ord }_ord ^thin \int_op

The \hphantom{...} acts as subformula. As can be seen, there is an additional thin space (\thinmuskip) between \hphantom{...} and the following operator.

This space can be canceled by \!:

& \hphantom{{}={}} \! \int

Macro \! negates the thin space, it is defined in LaTeX as:

\def\!{\mskip-\thinmuskip}

An alternative is to remove the empty subformulas in \hphantom and convert the \hphantom to a rel atom as a visible = would have been:

& \mathrel{\hphantom{=}} \int

Analysis:

{}_ord ^thick { =_rel }rel ^thick \intop

Example file with increased \thinmuskip to show the spacing problem more clearly:

\documentclass[a5paper]{article}

\usepackage{amsmath}

\begin{document}

\thinmuskip=40mu

\begin{align}
 \texttt{\detokenize{& = \int}}\colon
 & = \int \\
 \texttt{\detokenize{& \hphantom{{}={}} \int}}\colon
 & \hphantom{{}={}} \int \\
 \texttt{\detokenize{& \hphantom{{}={}} \! \int}}\colon
 & \hphantom{{}={}} \! \int \\
 \texttt{\detokenize{& \mathrel{\hphantom{=}} \int}}\colon
 & \mathrel{\hphantom{=}} \int
\end{align}
\end{document}

score 6 · Accepted Answer · answered May 28 '15 at 05:30

6

An alternative approach would be to set the spacing based on the fact that = is a relation. Therefore, use \mathrel{\phantom{=}} (\hphantom is not really needed):

enter image description here

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\begin{align}
   a & \mathrel{\hphantom{=}} \int \\
   b & = \int \\
   c & = F
\end{align}

\end{document}

answered May 28 '15 at 05:30

Werner

603,163

Thanks, that seems like a good option. I thought that \hphantom{{}={}} should effectively be telling LaTeX the same information as \mathrel, but I guess not. – bdforbes May 28 '15 at 06:02

Slight misalignment using hphantom equals sign in align

2 Answers2