[Tex/LaTex] How to use semibold italic cmbright for bold math symbols

Question

I want to make my bold math symbols in a document using cmbright semibold and italic. How can I do this? \usepackage{bm} is close, but uses bold instead of semibold.

Here is a minimal working example:

\documentclass{article}
\usepackage{cmbright}
\usepackage{bm}
\begin{document}
     \[\bm{f} = e\left(\bm{E} + \bm{v} \wedge \bm{B}\right)\]
\end{document}

I am trying to replace the bold italic letters in the equation with semibold italic letters.

Thanks.

Answer 1

I was formulating a response to your similar question about concmath when you deleted it, so I will also post that response here:

---

CMBRIGHT

While not recommended for general use (I'm sure the font police will issue citations after I post), one can (for an odd letter here or there) fake a bold font with a kerned overstrike.

Here, I do a .2pt horizontal and a .1pt vertical kern for the overstrike, hopefully reaching a font between the normal and the bold italic version. The three versions are the normal math font, the overstrike font, and the bold-italic font.

\documentclass{article}
\usepackage{cmbright}
\usepackage{bm}
\usepackage{scalerel}
\def\sbfCMB#1{\ThisStyle{\setbox0=\hbox{$\SavedStyle#1$}%
  #1\kern.2pt\kern-\wd0%
  \raisebox{.1pt}{$\SavedStyle#1$}}}
\begin{document}
     \[f = e\left(E + v \wedge B\right)\]
     \[\sbfCMB{f} = e\left(\sbfCMB{E} + \sbfCMB{v} \wedge \sbfCMB{B}\right)\]
     \[\bm{f} = e\left(\bm{E} + \bm{v} \wedge \bm{B}\right)\]
\end{document}

---

CONCMATH (simulating math bold font)

In this MWE, the left example is concmath, the middle is a kerned triplestrike of the mathrm concmath (using \bfC), while the right example is the default bold font (Computer Modern).

\documentclass{article}
\usepackage{concmath}
\usepackage{scalerel}
\def\bfC#1{\ThisStyle{\setbox0=\hbox{$\SavedStyle\mathrm{#1}$}%
  \mathrm{#1}\kern.15pt\kern-\wd0%
  \mathrm{#1}\kern.15pt\kern-\wd0%
  \mathrm{#1}}}
\begin{document}
$A = BC, \bfC{A}=\bfC{B}\bfC{C}, \mathbf{A}=\mathbf{B}\mathbf{C}$\par
$\scriptstyle 
 A = BC, \bfC{A}=\bfC{B}\bfC{C}, \mathbf{A}=\mathbf{B}\mathbf{C}$\par
$\scriptscriptstyle 
 A = BC, \bfC{A}=\bfC{B}\bfC{C}, \mathbf{A}=\mathbf{B}\mathbf{C}$\par
\end{document}