我在使用命令制作目录时遇到问题。这\tableofcontents是我的长乳胶文件。
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\tableofcontents
\newpage
\section*{\Huge Chapter 1}
\section*{\Large Symmetries in Classical Physics}
\subsection*{Introduction}
Symmetries plays an important role in the fields of phsyics.
\subsection{Mechanics of many particle System}
Let us consider a system of more than 2 particles such that
\begin{itemize}
\item The external forces applied on the particles be say $F^{(ext)}_{i}$ and
\item Forces applied by the system i.e. the interaction force be $F^{j}_{i}$
\end{itemize}
$\therefore$ From Newtons 2nd Law of motion, we can write\\
\begin{equation}
m_ia_i=\dot{p_i}=F^{(ext)}_{i}+\sum_{i \ne j}F^{j}_{i}, i=1,2,...N
\end{equation}
Where 'i' stands for denoting the particle and\\
$F_{i}^{ext}$ stands for external force acting on $i^{th}$ particle and\\
$F^{j}_{i}$ is the interaction force acting on the $i^{th}$ particle due to all other paricle (i.e j particle).\\
So $\therefore$ for N particles one can write it as \\
\begin{equation*}
\sum_{i \ne j}^{N}F^{j}_{i} ,where F_{i}^{i}=0.\end{equation*}
From Newtons 3rd law,one can write forces due to internal/interaction force as\\
\begin{equation}
F_{i}^{j}=-F_{j}^{i}.
\end{equation}
\section*{\large \boldmath Important points to be noted:}
\begin{itemize}
\item Interacyion force is only possible iff there are more than on particles and bheave as an central force.
\item Ad they occur in pair, forces among themselves are neglected.
\end{itemize}
For a closed system,where one knows that linear momentum is conserved i.e.external force acting on the system is zero.\\
$\implies$ A closed system can be reffered to as a inertial frame.\\
$\therefore$ From eq.(1) we have,\\
\begin{equation*}
\sum_i\dot{p_i}=\frac{d^2}{ dt^2}\sum_{i} m_{i}r_{i}=\sum_iF^{(ext)}_{i}+\sum_{i \ne j}F^{j}_{i}
\end{equation*}
\begin{equation*}
=\sum_{i}F_{i}^{ext}
\end{equation*} where we have used eqn.(2).
\newpage
This is because \\
As \begin{equation*}
\sum_{i \ne j}F^{j}_{i}=\sum_{i \ne j}F^{i}_{j}
\end{equation*}
\begin{equation*}
=\frac{1}{2}[\sum_{i \ne j}F^{j}_{i}+\sum_{i \ne j}F^{i}_{j}]=0 \text{(used dummy variable conncept)}
\end{equation*}
\begin{equation}
\implies\sum_{i}\dot{p_i} = F^{(ext)}_{i}
\end{equation}
To understand the physical significance of eqn.(1),using the concept of center of mass for many particles we have,\\
\begin{center}
\begin{equation}
\boldmath {R}=\frac{\sum_{i} m_{i}r_{i}}{\sum_{i}m_{i}}
\end{equation}
\end{center}
where M=$\sum_{i} m_{i}$ is the total mass of the system.\\
$\therefore$ eqn.(3) become,\\
\begin{equation}
M\frac{d^2R}{dt^2} = \sum_{i}F_{i}^{ext}=F^{ext}
\end{equation}
\begin{equation*}\textbf{P}=\sum_{i} m_{i}r_{i}=\frac{d}{dt}\sum_{i} m_{i}r_{i}=M\dot{\textbf{R}}
\end{equation*}
\section*{Conservation Laws}
\subsection{Linear momentum}
Frome eqn.(5) we have\\
\begin{equation*}\textbf{P}=\sum_{i} m_{i}r_{i}=\frac{d}{dt}\sum_{i} m_{i}r_{i}=M\dot{\textbf{R}}
\end{equation*}
\textbf{Important points}
\begin{itemize}
\item Applying the external force on the entire mass of the system, the C.O.M move along too.
\item If $F^{ext}$=0,then the total linear momentum is conserved.\\
i.e $\frac{d\textbf{P}}{dt}$=0 $\implies$ $\textbf{P}$=constant of motion.\\
$\implies$ C.O.M moves with cobstant/uniform velocity without the extrnal force.\\
\end{itemize}
We can relate this scenario by the radioactive decay nucleus disintegrating ejects different particles which move in different direction in such a way that C.O.M moves along with constant velocity after disnitegration.\\
\subsection{Conservation of angular momentum}
Angular momentum for the system of particles is written as:-\\
\begin{equation*}
\textbf{L}=\sum_{i}r_{i} \times p_{i}
\end{equation*}
\begin{equation*}
\dot{\textbf{L}}=\sum_{i}(r_{i} \times \dot{p_{i}}+\dot{r_{i}} \times p_{i})
\end{equation*}
\begin{equation*}
\dot{\textbf{L}}=\sum_{i}r_{i} \times \dot{p_{i}} (\because \dot{r_{i}} \times p_{i}=0)
\end{equation*}
\begin{equation*}
\dot{\textbf{L}}=r_{i} \times(\sum_{i}F^(ext)_i)+\sum_{i \ne j} F^{j}_{i}
\end{equation*}
\begin{equation}
\frac{d\dot{L}}{dt}=\sum_{i}r_{i} \times F^{(ext)}_{i}+\sum_{i \ne j}r_{i} \times F^{j}_{i}
\end{equation}
where 1st term represnts total torque due to external forces and 2nd term represents total torque due to internal forces.\\
Now one can easily prove that 2nd term in eqn.(6) is zero.\\
\textbf{Proof}:-Let $N_{i}$ be the sum of all internal torque acting on the system i.e.,\\
\begin{equation*} N_{i}=\sum_{i\ne j} r_{i} \times F_{i}^{j}=\sum_{i\ne j} r_{j} \times F_{i}^{j} \text{()used dummy varialble concept} \end{equation*}
\begin{equation*}
N_{i}=\frac{1}{2}\sum_{i \ne j} [ r_{i} \times F_{i}^{j}+ r_{j} \times F_{i}^{j}]
\end{equation*}
\begin{equation*}
\text{Now as} F_{i}^{j}= -F_{j}^{i}
\end{equation*}
\begin{equation*}
N_{i}=\frac{1}{2}\sum_{i \ne j} (r_{i}-r_{j}) \times F_{i}^{j}
\end{equation*}
\begin{equation*}
r_{ij}=r_{i}-r_{j}
\end{equation*}
\begin{equation*}
\implies N_{int}=\frac{1}{2} \times r_{ij} \times F^{j}_{i}
\end{equation*}
From the figure below one can see that $r_{ij} \times F^{j}_{i}$=0; $\because$ $r_{ij}$ of particle $i^{th}$ and $j^{th}$ are parallel to the line of action of $F^{j}_{i}$.
\begin{figure}[H]
\centering
\includegraphics[width=1.2\linewidth]{IMG_20200426_202229}
\caption*{Representation of $i_{th}$ and $j_{th}$ particles }
\label{fig:img20200426202229}
\end{figure}
\begin{equation*}
\implies\bf{ \frac{dL}{dt}=\sum_{i} r_{i} \times F_{i}^{ext}=N^{ext}}=\text{Sum of external torques}
\end{equation*}
Now if $N^{ext}$=0 $\implies$ \text{\textbf{'L'} is aconstant of motion and thus angular momentum is conserved}.
\subsection{Conservation of Energy}
\end{document}
据我所知，section 命令会自动在目录中添加标题，但\section*{}我看到它不包括那些，而且我在许多页面中制作了部分，使其在目录中变得粗体和大，它也变大了但我希望手动对其进行修改。对此的任何人都可以提供帮助（而且当我在 pdf 中预览它时，我会看到我的整个书面文本，并且喜欢居中并且不会占用整个页面）。