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脚注不跨页

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脚注不跨页

我有一个很长的脚注，其中包含几个方程式（来源如下）。尽管如此，LaTeX 还是拒绝将脚注拆分到一页，\interfootnotelinepenalty=0结果在脚注文本之前创建了一个几乎空白的页面。我发现如果我不这样做，问题就会消失usepackage{txfonts}。

这个答案演示如何强制分页在脚注中的某个特定位置，是的，在我的示例中可以这样。但是，我希望 LaTeX 能够自动合理地断开脚注，就像在不使用 时一样usepackage{txfonts}。

如何实现这一点？

% source code (edited after first answer attempt didn't solve problem)
\documentclass[12pt,fleqn]{article}
\usepackage{amsmath,amssymb,graphics}
\usepackage{txfonts}
\usepackage{mathtools}
\usepackage{empheq}

\interfootnotelinepenalty=0

\newcommand*{\p}         {\partial}
\renewcommand*{\vec}[1]  {\boldsymbol{#1}}
\newcommand*{\pdiff}[2]  {\frac{\p{#1}}{\p{#2}}}

\begin{document}
\section{A}
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This just some text preceding the footnote\footnote
{It is an instructive but non-trivial exercise to prove this result directly from the definition and without Euler's theorem. Taking the derivative with respect to the position of particle $i$ yields its force:
\[
    \vec{F}_i = - \pdiff{U}{\vec{r}_i} =
    - \frac{G}{2}\sum_{j,k} \frac{m_{\!j}\,m_k}{|\vec{r}_{\!j}-\vec{r}_k|^2}
    \frac{\vec{r}_{\!j}-\vec{r}_k}{|\vec{r}_{\!j}-\vec{r}_k|}
    (\delta_{i\!j}-\delta_{ik})
\]
where we have used $\p|\vec{r}|/\p\vec{r}=\vec{r}/|\vec{r}|$. Using the defining property of the Kronecker delta gives
\[
    \vec{F}_i = 
    - \frac{G}{2}
    \left[\sum_{k} \frac{m_{i}\,m_k}{|\vec{r}_{i}-\vec{r}_k|^2}
                   \frac{\vec{r}_{i}-\vec{r}_k}{|\vec{r}_{i}-\vec{r}_k|} -
          \sum_{j} \frac{m_{\!j}\,m_i}{|\vec{r}_{\!j}-\vec{r}_i|^2}
                   \frac{\vec{r}_{\!j}-\vec{r}_i}{|\vec{r}_{\!j}-\vec{r}_i|}
    \right].
\]
Since the sum index is just a dummy variable, we see that the two terms are identical,
i.e.
\[
    \vec{F}_i = - G \sum_{k} \frac{m_{i}\,m_k}{|\vec{r}_{i}-\vec{r}_k|^2}
                          \frac{\vec{r}_{i}-\vec{r}_k}{|\vec{r}_{i}-\vec{r}_k|}.
\]
We can now compute the virial of the system:
\[
    \sum_i\vec{F}_i\cdot\vec{r}_i =
    -G\sum_{i,k}
    \frac{m_{i}\,m_k}{|\vec{r}_{i}-\vec{r}_k|^2}
    \frac{\vec{r}_{i}-\vec{r}_k}{|\vec{r}_{i}-\vec{r}_k|}\cdot\vec{r}_i.
\]
Consider the same expression with the summation indices $i$ and $k$ swapped
\[
    \sum_i\vec{F}_i\cdot\vec{r}_i =
    -G\sum_{i,k}
    \frac{m_k\,m_i}{|\vec{r}_k-\vec{r}_i|^2}
    \frac{\vec{r}_k-\vec{r}_i}{|\vec{r}_k-\vec{r}_i|}\cdot\vec{r}_k 
    =  G\sum_{i,k}  \label{eq:U:grav:2}
    \frac{m_i\,m_k}{|\vec{r}_{i}-\vec{r}_k|^2}
    \frac{\vec{r}_i-\vec{r}_k}{|\vec{r}_i-\vec{r}_k|}\cdot\vec{r}_k,
\]
where the second equality follows from $\vec{r}_i-\vec{r}_k = -(\vec{r}_k-\vec{r}_i)$.
Adding this to the previous form,
\[
    \sum_i\vec{F}_i\cdot\vec{r}_i =
    - \frac{G}{2} \sum_{i,k}\frac{m_{i}\,m_k}{|\vec{r}_{i}-\vec{r}_k|^2}
    \frac{\vec{r}_{i}-\vec{r}_k}{|\vec{r}_{i}-\vec{r}_k|}\cdot(\vec{r}_i-\vec{r}_k)
    = -\frac{G}{2}
    \sum_{i,k}\frac{m_{i}\,m_k}{|\vec{r}_{i}-\vec{r}_k|}
    = U.
\]
}
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\end{document}

抱歉，我没能创建一个更简短的例子。

答案1

两条建议/意见：

  • 为了让 LaTeX 能够在长脚注中创建分页符，您必须做的主要事情是停止加载包fleqn。另外：由于您指定了文档类选项fleqn（将传递给包amsmath），因此无论如何都没有理由加载过时的fleqn包。

  • （可选）不要加载几乎过时的txfonts包，而要考虑加载较新的newtxtext包newtxmath。您会发现较新的包会生成间距更好的数学表达式。

答案2

LaTeX 不会拒绝拆分脚注：如果您稍加修改，例如添加 1-3 行文本，它就会起作用。此外，这\section本身并不是真正的问题，但它会改变文本的长度，从而改变分页。

主要问题是您在脚注中放入了太多大对象。在某些情况下（并非所有情况！），您可以通过在显示数学之前留出一个停顿来改善这种情况，但在我看来，最好避免使用这样的脚注——将证明放在附录中。

\usepackage{xpatch}

\makeatletter
\patchcmd\@footnotetext{\footnotesize}{\footnotesize \predisplaypenalty=-100 }{}{\fail}
\makeatother

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