problem with hspace

Question

I noticed now a space problem with hspace, especially the command:

\newcommand{\Div}[1]{\displaystyle \boldsymbol{\nabla}\hspace{-.1em}\cdot\hspace{-.1em} #1}

how could it be solved, keeping the same formalism?

external note :

%<*notes012>  
\footnote{ricordando che se $\Div{u}=0$ allora $\barbII{\tau}=-\textbf{p}\I+2\mu\barbII{e}$}
%</notes012>

---

L'equazione  costitutiva per fluidi newtoniani \eqref{eq062}, può essere espressa nella sua forma per componenti\loadnote{012}:

\begin{equation}\label{eq063}
\begin{aligned}
    \tau_{ij}=-\textbf{p}\delta_{ij}+2\mu\left( \eij-\Fra{1}{3}e_{kk}\delta_{ij}\right) 
    \text{dove $\delta_{ij}$ = }~\left\{\begin{array}{rcl}
    1&\rightarrow& \text{se i = j;}\\
    0&\rightarrow& \text{se i $\neq$ j};\\
    \end{array}\right.\\
\end{aligned}
\end{equation}

\begin{tikzpicture}[right node/.style={rectangle,draw}]
 node{%
        $\begin{aligned}
            \tikzmarknode{ex1}e_{kk}\tikzmarknode{ex2}=tr(\barbII{e})=\tikzmarknode{ex3}\Div{u}
        \end{aligned}$};

    \draw[->,>=latex]([shift={(-70pt,-5pt)}]pic cs:ex1-ex2) |- ++(80pt,-25pt) node[right,text width=6cm] 
    { Considera la somma degli elementi della diagonale di $\barbII{e}$ e non coinvolge gli elementi ij;};
    \draw[->,>=latex]([shift={(-8pt,-15pt)}]pic cs:ex3) |- ++(16pt,-52pt) node[right,text width=6cm] 
    {Velocità di Deformazione Volumetrica (Comprimibiltà);};
\end{tikzpicture}

the \nabla}\hspace{-.1em}\cdot\hspace{-.1em} has different distances. when it is inserted in the command \begin{tikzpicture}.... the distances are not respected

Answer 1

I guess you want the negative space in order to reduce the distance between the centered dot and the symbols at both sides.

Using the em unit is wrong to begin with, because it depends on the text font current at the time the formula started. In a tikzpicture the font might be \nullfont for which the em unit corresponds to zero and, in any case, it is not generally predictable.

Beware also of \displaystyle which serves no real purpose for typesetting “nabla+centered dot+variable” and will make damage to the rest of the formula.

My choice would be, loading also the bm package after amsmath (the package provides for a better replacement of \boldsymbol called \bm, but allows \boldsymbol as a synonym for \bm, so your code need not be changed):

\newcommand{\Div}{\bm{\nabla}\!\cdot\!}

The \! command inserts some negative space using “math units”, which are independent on the current text font. There's no need to use an argument; however, \Div u and \Div{u} would do the same.

You should also avoid \textbf in formulas for math symbols: the correct command is \mathbf as this will not be affected by font parameters (shape, for instance) current at the time the formula started out.

By the way, input such as

\text{dove $\delta_{ij}$ = }~\left\{\begin{array}{rcl}
    1&\rightarrow& \text{se i = j;}\\
    0&\rightarrow& \text{se i $\neq$ j};\\

is incorrect under many respects. It should be

\text{dove }\delta_{ij} = 
\begin{cases}
  1 & \text{se $i = j$;} \\
  0 & \text{se $i \neq j$;}
\end{cases}

I've never seen an arrow between the value for the case and the condition; however, this is not the main point. The letters “i” and “j” denote the very same object as in the subscript to \delta, so they must be typeset in the same way, that is, math italic.

As a stylistic remark, the “Kronecker delta” is so widespread a symbol that it should not be explained as part of a more important explanation: this is very distracting. Define the symbol beforehand and use it without explicit mention.

A grammatical note: Velocit\'a typesets an acute accent whereas it should be a grave accent. Thus Velocit\`a. Or, why not exploiting modern operating systems and type Velocità?

Answer 2

This may not be an answer (but given that you do not provide us with some compilable code, this is the best I can do at this point). This version

• avoids nesting tikzpictures and
• uses \tikzmarknode in an arguably more appropriate way.

Apart from that, I guessed what your macros are doing. And I removed redundancies (why are you adding aligned here, you were not using it).

\documentclass[fleqn]{article}
\usepackage{amsmath,bbm}
\usepackage{tikz}
\usetikzlibrary{tikzmark}
\newcommand{\Div}[1]{\displaystyle \boldsymbol{\nabla}\!\cdot\! #1}
\newcommand{\eij}[0]{\ensuremath{e_{ij}}}
\newcommand{\Fra}[2]{\frac{#1}{#2}}
\newcommand{\barbII}[1]{\underset{=}{#1}}
\DeclareMathOperator{\tr}{tr}
\begin{document}

\begin{align}
    \tau_{ij}&=-\boldsymbol{p}\delta_{ij}+2\mu\left( \eij-\Fra{1}{3}e_{kk}\delta_{ij}\right) 
    \text{dove $\delta_{ij}$ = }~\begin{cases}
    1&\rightarrow \text{se}~i = j;\\
    0&\rightarrow \text{se}~i\neq j;
    \end{cases}\label{eq063}\\
    \tikzmarknode{ex1}{e_{kk}}&=\tr(\barbII{e})=\Div{\tikzmarknode{ex3}{u}} 
\end{align}
\bigskip\bigskip\bigskip\bigskip
\begin{tikzpicture}[right node/.style={rectangle,draw},overlay,remember picture]
    \draw[-latex]([yshift=-1pt]ex1.south) |- ++(86pt,-25pt) 
    node[right,text width=6cm] (cons)
    { Considera la somma degli elementi della diagonale di $\barbII{e}$ e non coinvolge gli elementi ij;};
    \draw[-latex,preaction={draw,white,ultra thick,-}]([yshift=-1pt]ex3.south) -- ++(0,-60pt) coordinate(aux)
    -- (aux -| cons.west) node[right,text width=6cm] 
    {Velocit\'a di Deformazione Volumetrica (Comprimibilt\'a);};
\end{tikzpicture}

blah blah\footnote{ricordando che se $\Div{u}=0$ allora $\barbII{\tau}=-\boldsymbol{p}
\boldsymbol{I}+2\mu\barbII{e}$} 
\end{document}