mathpazo + siunitx: π turns into ß

Question

When I use mathpazo and siunitx (version 2.3h) together, if I use \pi inside the numerical argument to \SI, it will show up in the output as "ß".

\documentclass{article}
\usepackage{mathpazo, siunitx}
\begin{document}
    $\pi + \SI{\pi}{\ohm}$
\end{document}

Produces:

But when I don't use mathpazo:

Why does this happen and how can I fix it?

Answer 1

As others have observed, the issue here is the way the mathpazo treats \pi. I would go with

\documentclass{article}
\usepackage{mathpazo, siunitx}
\protected\def\numpi{\text{\ensuremath{\pi}}}
\sisetup{input-symbols = \numpi}
\begin{document}
    $\pi + \SI{\numpi}{\ohm}$
\end{document}

The idea here is that this approach forces the use of the 'standard' font for \pi in all cases (it's safe in both math and text mode). There are reasons that siunitx uses \mathrm rather than \mathnormal as the standard math mode font: try

\documentclass{article}
\begin{document}
    $\mathnormal{123}$
\end{document}

to see why!

Answer 2

The following rectifies your problem:

• \pi without siunitx or mathpazo is defined as \mathchar"119;
• \pi under only siunitx remains the same (\mathchar"119);
• \pi under only mathpazo is defined as \mathchar"7119.

Since siunitx does not touch the definition of \pi and the load order does not correct for this, redefining \pi works:

\documentclass{article}
\usepackage{siunitx,mathpazo}
\begin{document}
  $\pi + \SI{\pi}{\ohm}$ \par
  \renewcommand{\pi}{\mathchar"119}% Revert to original mathchar definition for \pi
  $\pi + \SI{\pi}{\ohm}$ \par
\end{document}

Answer 3

The problem is that mathpazo defines \pi as \mathalpha, that is it adds "7000 to its code and that siunitx uses by default \mathrm in the arguments of \SI:

\sisetup{number-math-rm=\mathnormal}

would be wrong; use

\SI[number-math-rm=\mathnormal]{\pi}{\ohm}

Answer 4

Sorry for my previous wrong answer. Here is a tested code. It looks more concise.

\documentclass{article}
\usepackage{mathpazo, siunitx}
\begin{document}
    $\pi + \SI[mode=text]{\pi}{\ohm}$

    $\pi + \SI{\pi}{\ohm}$

    $\pi + \SI[mode=math]{\pi}{\ohm}$
\end{document}

Note: The code will not work if you replace the '\pi' inside the numerical argument by other non-numeric symbols like 'p'.