# long footnote not split over two pages

I have a long footnote which LaTeX refuses to split over two pages, but rather leaves the previous page half empty (so that the text with the footnote appears on a new page with enough space for the long footnote to follow). How can I prevent that so that there is no half empty page and the footnote broken over two pages?

It's not easy to reproduce this with little code. Here is a snippet from my actual file

\documentclass[12pt,fleqn]{article}
\usepackage{amsmath,amssymb,graphics,fleqn,txfonts}
\usepackage{mathtools}
\usepackage{empheq}

\newcommand*{\p}         {\partial}
\newcommand*{\B}      {\boldsymbol{#1}}
\newcommand*{\pdiff}  {\frac{\p{#1}}{\p{#2}}}

\begin{document}
\subsubsection{Application to self-gravitating systems}
The total gravitational potential energy of a self-gravitating system of $n$ point-like
particles with masses $m_i$ and positions $\B{r}_i$ is
\begin{equation} \label{eq:U:selfgrav}
U = -\frac{G}{2}\sum_{j,k} \frac{m_{\!j}\,m_k}{|\B{r}_{\!j}-\B{r}_k|}.
\end{equation}
Here, the factor $\tfrac{1}{2}$ accounts for the double counting of each particle
pair in the double sum over $j$ and $k$. If we multiply each $\B{r}_i$ with the
same constant $\alpha$, the gravitational energy is reduced by $\alpha^{-1}$. In other
words, $U$ is a homogeneous function of degree $-1$ and we can apply Euler's
theorem(\ref{eq:homo:theorem}) for homogeneous functions to obtain for a
self-gravitating system%
%
% begin of footnote
%
\footnote{It is an instructive but non-trivial exercise to prove
this result directly from the definition~(\ref{eq:U:selfgrav}). Taking the derivative
with respect to the position of particle $i$ yields its force:
$\B{F}_i = - \pdiff{U}{\B{r}_i} = - \frac{G}{2}\sum_{j,k} \frac{m_{\!j}\,m_k}{|\B{r}_{\!j}-\B{r}_k|^2} \frac{\B{r}_{\!j}-\B{r}_k}{|\B{r}_{\!j}-\B{r}_k|} (\delta_{i\!j}-\delta_{ik})$
where we have used $\p|\B{r}|/\p\B{r}=\B{r}/|\B{r}|$. Using the defining
property~(\ref{eq:delta}) of the Kronecker delta gives
$\B{F}_i = - \frac{G}{2} \left[\sum_{k} \frac{m_{i}\,m_k}{|\B{r}_{i}-\B{r}_k|^2} \frac{\B{r}_{i}-\B{r}_k}{|\B{r}_{i}-\B{r}_k|} - \sum_{j} \frac{m_{\!j}\,m_i}{|\B{r}_{\!j}-\B{r}_i|^2} \frac{\B{r}_{\!j}-\B{r}_i}{|\B{r}_{\!j}-\B{r}_i|} \right].$
Since the sum index is just a dummy variable, we see that the two terms are identical,
i.e.
$\B{F}_i = - G \sum_{k} \frac{m_{i}\,m_k}{|\B{r}_{i}-\B{r}_k|^2} \frac{\B{r}_{i}-\B{r}_k}{|\B{r}_{i}-\B{r}_k|}.$
We can now compute the virial of the system:
$\sum_i\B{F}_i\cdot\B{r}_i = -G\sum_{i,k} \frac{m_{i}\,m_k}{|\B{r}_{i}-\B{r}_k|^2} \frac{\B{r}_{i}-\B{r}_k}{|\B{r}_{i}-\B{r}_k|}\cdot\B{r}_i.$
Consider the same expression with the summation indices $i$ and $k$ swapped
$\sum_i\B{F}_i\cdot\B{r}_i = -G\sum_{i,k} \frac{m_k\,m_i}{|\B{r}_k-\B{r}_i|^2} \frac{\B{r}_k-\B{r}_i}{|\B{r}_k-\B{r}_i|}\cdot\B{r}_k = G\sum_{i,k} \label{eq:U:grav:2} \frac{m_i\,m_k}{|\B{r}_{i}-\B{r}_k|^2} \frac{\B{r}_i-\B{r}_k}{|\B{r}_i-\B{r}_k|}\cdot\B{r}_k,$
where the second equality follows from $\B{r}_i-\B{r}_k = -(\B{r}_k-\B{r}_i)$.
Adding this with the previous form,
$\sum_i\B{F}_i\cdot\B{r}_i = - \frac{G}{2} \sum_{i,k}\frac{m_{i}\,m_k}{|\B{r}_{i}-\B{r}_k|^2} \frac{\B{r}_{i}-\B{r}_k}{|\B{r}_{i}-\B{r}_k|}\cdot(\B{r}_i-\B{r}_k) = -\frac{G}{2} \sum_{i,k}\frac{m_{i}\,m_k}{|\B{r}_{i}-\B{r}_k|},$
which is just the total potential energy $U$.}
%
% end of footnote
%
Thus, for a self-gravitating system
\begin{equation}
\boxed{
\sum_i\pdiff{U}{\B{r}_i}\cdot\B{r}_i = - U.
}
\end{equation}
Hence, for gravitational systems, the virial theorem states that
\begin{equation} \label{eq:VT:grav}
\boxed{
2 \langle T\rangle + \langle U\rangle = 0.
}
\end{equation}
Since the total energy $E=T+U$, this also implies that for a self-gravitating system in
virial equilibrium
\begin{equation}
E = -T = \tfrac{1}{2}U.
\end{equation}
Thus, self-gravitating systems have a negative heat capacity: when reducing their
energy by cooling (for example by radiation), they become hotter, since $T=-E$. In
fact, their kinetic energy increases by the same amount as that lost.

The gravitational energy of a self-gravitating system with mass $M$ and size $R$ is
\begin{equation} \label{eq:U:GMMR}
U = -f\frac{GM^2}{2R},
\end{equation}
where $f$ is a geometric factor of order unity that depends on the details of the mass
distribution. Therefore, a young star which has not yet ignited fusion must shrink to
compensate the loss of energy by radiation off its surface. Since $T=-U/2$, this makes
it hotter.

A cluster of galaxies with observed line-of-sight velocity dispersion $\sigma$ has
kinetic energy
\begin{equation}
T = \tfrac{3}{2}M\sigma^2.
\end{equation}
Inserting this and equation~(\ref{eq:U:GMMR}) into the virial theorem and solving for
$M$, we get
\begin{equation}
M = \frac{3\sigma^2R}{fG}.
\end{equation}
Using this relation, Fritz Zwicky uncovered dark matter in the Virgo cluster of
galaxies in 1934.

\end{document}

• A bit difficult to see the issue with your minimal code.... – karlkoeller Feb 10 '15 at 9:57
• Please make your code snippet compilable for us, showing the problem you have. Then we can help you ... – Mensch Feb 10 '15 at 10:16
• Can't you provide us with a MWE that let's us actually reproduce the issue? – cgnieder Feb 10 '15 at 10:17
• haven't tried to compile this, but it looks like there are only two places in the footnote -- at the paragraph breaks -- where a page break would be entirely "acceptable". (penalties discourage breaks in paragraphs containing fewer than three lines; breaks before display math are prohibited; there's also an \interfootnotelinepenalty. values shown in latex.ltx.) it's a difficult situation. try putting in a couple of blank lines after displays (in places where you think the length might be appropriate for a break) or try resetting penalties and see what happens. – barbara beeton Feb 10 '15 at 16:11

I played a little bit with your code and with a quick and dirty trick I got only two pages and a split in the footnote. I added some blank lines and played with the position of a \newpage in the footnote ... I also commented package fleqn to get rid of the message.

Please try the following MWE (and see the line marked with % ============):

\documentclass[12pt,fleqn]{article}
\usepackage{amsmath,amssymb,graphics,txfonts} % ====
%\usepackage{fleqn}                           % ====
\usepackage{mathtools}
\usepackage{empheq}

\newcommand*{\p}         {\partial}
\newcommand*{\B}      {\boldsymbol{#1}}
\newcommand*{\pdiff}  {\frac{\p{#1}}{\p{#2}}}

%\interfootnotelinepenalty=0
%\sloppy

\begin{document}
\subsubsection{Application to self-gravitating systems}
The total gravitational potential energy of a self-gravitating system of $n$ point-like
particles with masses $m_i$ and positions $\B{r}_i$ is
\begin{equation} \label{eq:U:selfgrav}
U = -\frac{G}{2}\sum_{j,k} \frac{m_{\!j}\,m_k}{|\B{r}_{\!j}-\B{r}_k|}.
\end{equation}
Here, the factor $\tfrac{1}{2}$ accounts for the double counting of each particle
pair in the double sum over $j$ and $k$. If we multiply each $\B{r}_i$ with the
same constant $\alpha$, the gravitational energy is reduced by $\alpha^{-1}$. In other
words, $U$ is a homogeneous function of degree $-1$ and we can apply Euler's
theorem(\ref{eq:homo:theorem}) for homogeneous functions to obtain for a
self-gravitating system%
%
% begin of footnote
%
\footnote{It is an instructive but non-trivial exercise to prove
this result directly from the definition~(\ref{eq:U:selfgrav}).
Taking the derivative %\interfootnotelinepenalty=0
with respect to the position of particle $i$ yields its force:
$\B{F}_i = - \pdiff{U}{\B{r}_i} = - \frac{G}{2}\sum_{j,k} \frac{m_{\!j}\,m_k}{|\B{r}_{\!j}-\B{r}_k|^2} \frac{\B{r}_{\!j}-\B{r}_k}{|\B{r}_{\!j}-\B{r}_k|} (\delta_{i\!j}-\delta_{ik})$
where we have used $\p|\B{r}|/\p\B{r}=\B{r}/|\B{r}|$.

Using the defining property~(\ref{eq:delta}) of the Kronecker delta gives
$\B{F}_i = - \frac{G}{2} \left[\sum_{k} \frac{m_{i}\,m_k}{|\B{r}_{i}-\B{r}_k|^2} \frac{\B{r}_{i}-\B{r}_k}{|\B{r}_{i}-\B{r}_k|} - \sum_{j} \frac{m_{\!j}\,m_i}{|\B{r}_{\!j}-\B{r}_i|^2} \frac{\B{r}_{\!j}-\B{r}_i}{|\B{r}_{\!j}-\B{r}_i|} \right].$

Since the sum index is just a dummy variable, we see that the two terms
are identical, i.e.
$\B{F}_i = - G \sum_{k} \frac{m_{i}\,m_k}{|\B{r}_{i}-\B{r}_k|^2} \frac{\B{r}_{i}-\B{r}_k}{|\B{r}_{i}-\B{r}_k|}.$

We can now compute the virial of the system:
$\sum_i\B{F}_i\cdot\B{r}_i = -G\sum_{i,k} \frac{m_{i}\,m_k}{|\B{r}_{i}-\B{r}_k|^2} \frac{\B{r}_{i}-\B{r}_k}{|\B{r}_{i}-\B{r}_k|}\cdot\B{r}_i.$

Consider the same expression with the summation indices $i$ and $k$ swapped
$\sum_i\B{F}_i\cdot\B{r}_i = -G\sum_{i,k} \frac{m_k\,m_i}{|\B{r}_k-\B{r}_i|^2} \frac{\B{r}_k-\B{r}_i}{|\B{r}_k-\B{r}_i|}\cdot\B{r}_k = G\sum_{i,k} \label{eq:U:grav:2} \frac{m_i\,m_k}{|\B{r}_{i}-\B{r}_k|^2} \frac{\B{r}_i-\B{r}_k}{|\B{r}_i-\B{r}_k|}\cdot\B{r}_k,$
where the second equality follows from $\B{r}_i-\B{r}_k = -(\B{r}_k-\B{r}_i)$.
\newpage % =============================================================
Adding this with the previous form,
$\sum_i\B{F}_i\cdot\B{r}_i = - \frac{G}{2} \sum_{i,k}\frac{m_{i}\,m_k}{|\B{r}_{i}-\B{r}_k|^2} \frac{\B{r}_{i}-\B{r}_k}{|\B{r}_{i}-\B{r}_k|}\cdot(\B{r}_i-\B{r}_k) = -\frac{G}{2} \sum_{i,k}\frac{m_{i}\,m_k}{|\B{r}_{i}-\B{r}_k|},$
which is just the total potential energy $U$.}
%
% end of footnote
%
Thus, for a self-gravitating system
\begin{equation}
\boxed{
\sum_i\pdiff{U}{\B{r}_i}\cdot\B{r}_i = - U.
}
\end{equation}
Hence, for gravitational systems, the virial theorem states that
\begin{equation} \label{eq:VT:grav}
\boxed{
2 \langle T\rangle + \langle U\rangle = 0.
}
\end{equation}
Since the total energy $E=T+U$, this also implies that for a self-gravitating system in
virial equilibrium
\begin{equation}
E = -T = \tfrac{1}{2}U.
\end{equation}
Thus, self-gravitating systems have a negative heat capacity: when reducing their
energy by cooling (for example by radiation), they become hotter, since $T=-E$. In
fact, their kinetic energy increases by the same amount as that lost.

The gravitational energy of a self-gravitating system with mass $M$ and size $R$ is
\begin{equation} \label{eq:U:GMMR}
U = -f\frac{GM^2}{2R},
\end{equation}
where $f$ is a geometric factor of order unity that depends on the details of the mass
distribution. Therefore, a young star which has not yet ignited fusion must shrink to
compensate the loss of energy by radiation off its surface. Since $T=-U/2$, this makes
it hotter.

A cluster of galaxies with observed line-of-sight velocity dispersion $\sigma$ has
kinetic energy
\begin{equation}
T = \tfrac{3}{2}M\sigma^2.
\end{equation}
Inserting this and equation~(\ref{eq:U:GMMR}) into the virial theorem and solving for
$M$, we get
\begin{equation}
M = \frac{3\sigma^2R}{fG}.
\end{equation}
Using this relation, Fritz Zwicky uncovered dark matter in the Virgo cluster of
galaxies in 1934.

\end{document}


With the result for the first page: 