I have this simple code,
\documentclass[12pt]{book}
\usepackage{mathtools}
\usepackage{newtxtext}
\usepackage[lite]{mtpro2}
\begin{document}
$d\bar{p}$ and $dt$.
\end{document}
with this output,
I think it's a canonical space is due to the presence of the \bar{...} command? Sometimes to bring it back I insert a negative space \! to bring the p to the left near the d.
This little adjustment must always be done or I should create a particular macro. Which one could be always using the same command \bar? It gets a bit complicated for me in my book to change the original command \bar{...}. Thank you.
Using Steven Segletes' suggestion of \ooalign{$d\bar{#1}$\cr$d#1$}, one can see if the problem is universal across \bar. If the superposition produces a "thick" second letter, the kerning is not correct. (using newtxmath as Steven does not have mtpro2)
Unfortunately, that proves not to be the case, as some cases seem kerned correctly and others not:
\documentclass[12pt]{book}
\usepackage{mathtools}
\usepackage{newtxtext}
\usepackage{newtxmath}%instead of \usepackage[lite]{mtpro2}
\newcommand\testkern[1]{%
\ooalign{$d\bar{#1}$\cr$d#1$}}
\textwidth 1in
\begin{document}
\noindent\testkern{a}
\testkern{b}
\testkern{c}
\testkern{d}
\testkern{e}
\testkern{f}
\testkern{g}
\testkern{h}
\testkern{i}
\testkern{j}
\testkern{k}
\testkern{l}
\testkern{m}
\testkern{n}
\testkern{o}
\testkern{p}
\testkern{q}
\testkern{r}
\testkern{s}
\testkern{t}
\testkern{u}
\testkern{v}
\testkern{w}
\testkern{x}
\testkern{y}
\testkern{z}
\end{document}
Applying a universal -1mu kern to \bar
\newcommand\xbar{\mkern-1mu\bar}
\newcommand\testkern[1]{%
\ooalign{$d\xbar{#1}$\cr$d#1$}}
therefore, may fix some letter combinations, but will inevitably break others:








mtpro2, the space does not manifest. However, it does show, when, instead ofmtpro2, one usesnewtxmath. – Steven B. Segletes Apr 04 '19 at 10:18mtpro2because I like fonts. – Sebastiano Apr 04 '19 at 10:20\ooalign{$d\bar{p}$\cr$dp$}– Steven B. Segletes Apr 04 '19 at 10:21\barin that font, you could always do\let\svbar\bar\renewcommand\bar{\mkern-1mu\svbar}– Steven B. Segletes Apr 04 '19 at 10:24p's are "thick", that means the kerning is different between the two cases. If the superimposed p's look like one, then the kerning is correct. – Steven B. Segletes Apr 04 '19 at 10:26