20 \boldsymbol与\bm的区别以及出现`TeX capacity exceeded`的疑问

发布于 2024-10-15 22:50:33

转交流群友遇到的一个问题:
在article类中使用下面的代码可以正常编译:

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation*}
    \begin{aligned}
     & \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{1}{2}\dot{\boldsymbol{x}}\boldsymbol{^{\top } m}\dot{\boldsymbol{x}} +V(\boldsymbol{x})\right) =\frac{1}{2}\left(\ddot{\boldsymbol{x}}\boldsymbol{^{\top } m}\dot{\boldsymbol{x}} +\boldsymbol{\dot{\boldsymbol{x}}^{\top } m}\ddot{\boldsymbol{x}}\right) +\dot{\boldsymbol{x}}^{\top } \nabla V(\boldsymbol{x})\\
    = & \boldsymbol{\dot{\boldsymbol{x}}^{\top }}(\boldsymbol{m}\ddot{\boldsymbol{x}} +\nabla V(\boldsymbol{x})) =\boldsymbol{\dot{\boldsymbol{x}}^{\top } 0} =0,
    \end{aligned}
    \end{equation*}
\end{document}

但在elegantbook模板中使用则会出现下面的报错信息

\documentclass[lang=cn,newtx,10pt,scheme=chinese]{elegantbook}
\begin{document}
\begin{equation}
    \begin{aligned}
    & \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{1}{2}\dot{\boldsymbol{x}}\boldsymbol{^{\top } m}\dot{\boldsymbol{x}} +V(\boldsymbol{x})\right) =\frac{1}{2}\left(\ddot{\boldsymbol{x}}\boldsymbol{^{\top } m}\dot{\boldsymbol{x}} +\boldsymbol{\dot{\boldsymbol{x}}^{\top } m}\ddot{\boldsymbol{x}}\right) +\dot{\boldsymbol{x}}^{\top } \nabla V(\boldsymbol{x})\\
    = & \boldsymbol{\dot{\boldsymbol{x}}^{\top }}(\boldsymbol{m}\ddot{\boldsymbol{x}} +\nabla V(\boldsymbol{x})) =\boldsymbol{\dot{\boldsymbol{x}}^{\top } 0} =0,
    \end{aligned}
\end{equation}
\end{document}

报错信息如下，猜测是因为\boldsymbol命令的嵌套使用导致的...但是也很奇哉怪也

Runaway definition?
->
! TeX capacity exceeded, sorry [main memory size=5000000].
\bm@add ...mand #1}\xdef \bm@command {\the \toks@
                                                  }\endgroup \bm@expand
l.5 ...\boldsymbol{\dot{\boldsymbol{x}}^{\top } m}
                                                  \ddot{\boldsymbol{x}}\righ...

No pages of output.

群友S老师@u79794 提出使用\bm可以解决这一问题

% preamble
\usepackage{fixdif,derivative,physics2,bm}
\usephysicsmodule{ab}

\begin{equation*}
\begin{aligned}
  & \odv{}t \ab(\frac12 \dot{\bm x}+V(\bm x))
   =\frac12\ab(\ddot{\bm x}^\top\bm m\dot{\bm x}+\dot{\bm x}^\top\bm m+\ddot{\bm x})+\dot{\bm x}^\top\nabla V(\bm x)\\
  & =\dot{\bm x}^\top(\bm m\ddot{\bm x}+\nabla V(\bm x))=\dot{\bm x}^\top\bm 0=0,
\end{aligned}
\end{equation*}