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	\begin{document}

      \begin{center}
      {\large Classical Mechanics, Lecture 4 \\}  
      {\small January 29, 2008 \\}  
      {\small John Baez\\}  
      {\small Homework by Michael Maroun}  
      \end{center}

% naming format: Problem Set#.Problem#(within Set)

\section*{Conservation of Energy in the $n$-Body Problem}
Show that if Newton's $2^{nd}$ law ($F_i(t) = m_{i}\,\ddot{\q}_{i}(t)$) holds then energy is conserved:
\[\frac{d}{dt}\E(t)~=~0.\]

\section*{\it{Solution:}}
Below are the three equations for the total potential energy, the total kinetic energy, and the total energy of the system respectively:
\begin{eqnarray*}
\V(t) & = & \sum_{i=1}^{n}\V_i(t)~=~\sum_{i\,=\,1}^{n}\sum_{j\,\neq\,i}\V_{ij}(\mid\q_{i}(t)-\q_{j}(t)\mid) \\
\T(t) & = & \sum_{i\,=\,1}^{n}\frac{1}{2}\;m_{i}\,\dot{\q}_{i}(t)^{2}. \\
\E(t) & = &  \T(t)~+~\V(t).
\end{eqnarray*}
We can differentiate $\E(t)$ with respect to time explicitly.
\begin{eqnarray}
\frac{d\E(t)}{dt} & = & \frac{d}{dt}\left[ T(t)~+~V(t)\right] \nonumber \\
\frac{d\E(t)}{dt} & = & \frac{d}{dt}\left[\sum_{i\,=\,1}^{n}\frac{1}{2}\,m_{i}\,\dot{\q}_i(t)^2 + \sum_{i\,=\,1}^{n}\V_i(t)\right] \nonumber \\
& = & \sum_{i\,=\,1}^{n}\frac{1}{2}\,m_{i}\,\frac{d}{dt}\dot{\q}_{i}(t)^{2} + \sum_{i\,=\,1}^{n}\frac{d\V_i(t)}{dt} \nonumber \\
& = & \sum_{i\,=\,1}^{n}\,m_{i}\,\ddot{\q}_{i}(t)\,\dot{\q}_{i}(t) + \sum_{i\,=\,1}^{n}\frac{\partial\V_i(\mid\q_{i}(t)-\q_{j}(t)\mid)}{\partial\q_i}\,\dot{\q}_{i}(t) \nonumber \\
& = & \sum_{i\,=\,1}^{n}\,\dot{\q}_{i}(t)\left[\,m_{i}\,\ddot{\q}_{i}(t) + \frac{\partial\V_i(\mid\q_{i}(t)-\q_{j}(t)\mid)}{\partial\q_i}\right]
\label{1}
\end{eqnarray}
Because the potential $\V_{i}(\mid\q_i(t)-\q_j(t)\mid)$ depends only on $\mid\q_i(t)-\q_j(t)\mid$ we get:
\begin{equation}
\frac{\partial\V_i(\mid\q_{i}(t)-\q_{j}(t)\mid)}{\partial\q_i}\,~=~\sum_{j\,\neq\,i}\frac{\partial\V_{ij}(\mid\q_{i}(t)-\q_{j}(t)\mid)}{\partial\q_i}~=~\sum_{j\,\neq\,i}\V^\prime_{ij}\,\frac{\q_{i}(t)-\q_{j}(t)}{\mid\q_i(t)-\q_j(t)\mid}.
\label{2}
\end{equation}
Newton's $2^{nd}$ law here reads:
\begin{equation}
F_i(t)~=~m_{i}\,\ddot{\q}_i(t)~=~\sum_{j\,\neq\,i}F_{ij}(t)~=~\sum_{j\,\neq\,i}f_{ij}(\mid\q_i(t)-\q_j(t)\mid)\frac{\q_i(t)-\q_j(t)}{\mid\mid\q_i(t)-\q_j(t)\mid}.
\label{3}
\end{equation}
Since within the classical regime Newton's $2^{nd}$ law always holds, when we say ``Show that if Newton's $2^{nd}$ law holds...'', we mean given that the symmetric inter-particle interaction potential is derivable from a conservative force, i.e.
\begin{equation}
\V^\prime_{ij}~=~-f_{ij},
\label{4}
\end{equation}
implies that indeed the total energy of the system is conserved. Hence substituting equations \eqref{2} and \eqref{3} into \eqref{1} gives:
\begin{equation}
\frac{d\E(t)}{dt} = \sum_{i\,=\,1}^{n}\,\dot{\q}_{i}(t)\left[\sum_{j\,\neq\,i}f_{ij}(\mid\q_i(t)-\q_j(t)\mid)\frac{\q_i(t)-\q_j(t)}{(\mid\q_i(t)-\q_j(t)\mid)} + \sum_{j\,\neq\,i}\V^\prime_{ij}\,\frac{\q_{i}(t)-\q_{j}(t)}{\mid\q_i(t)-\q_j(t)\mid}\right].
\label{5}
\end{equation}
Now substituting \eqref{4} into \eqref{5}, we find the desired result:
\begin{center}
\fbox{
$\displaystyle\frac{d\E(t)}{dt}~=~\sum_{i\,=\,1}^{n}\,\dot{\q}_{i}(t)\sum_{j\,\neq\,i}\bigg[f_{ij}(\mid\q_i(t)-\q_j(t)\mid)-f_{ij}(\mid\q_i(t)-\q_j(t)
\mid)\bigg]\frac{\q_i(t)-\q_j(t)}{(\mid\q_i(t)-\q_j(t)\mid)}~=~0$}
\end{center}
We could have just as well arrived at the same valid conclusion by omitting \eqref{4} and noting that $\frac{\partial\V_ij}{\partial\q_i}~=~-\frac{\partial\V_ij}{\partial\q_j}$ because the quantity $\q_i(t)-\q_j(t)$ is anti-symmetric on interchange of $i$ and $j$. This is why $f_{ij}~=~f_{ji}$ but $F_{ij}~=~-F_{ji}$. This is of course what makes \eqref{4} true when the assumption of the existence of a conservative force interaction connected to a potential with the precise argument dependence of $\mid\q_i(t)-\q_j(t)\mid$ is made.

\section*{Conservation of Angular Momentum in the $n$-Body Problem:}
Show that $\frac{d}{dt}\J(t) = 0$ using Newton's $2^{nd}$ law and
\[F_{ij}(t)~=~f_{ij}(\mid\q_i(t)-\q_j(t)\mid)\frac{\q_i(t)-\q_j(t)}{\mid\q_i(t)-\q_j(t)\mid}\]
where $f_{ij}~=~f_{ji}$.

\section*{\it{Solution:}}
We start with the following:
\[\J(t)~=~\sum^{n}_{i=1}\J_i(t)~=~\sum^{n}_{i=1}m_i\q_i(t)\,\times~\dot{q}_i(t)\]
Differentiating the above equation gives:
\begin{flushleft}
\begin{eqnarray}
\frac{d\J(t)}{dt} & = & \frac{d}{dt}\left[\sum^{n}_{i=1}m_i\q_i(t)\,\times~\dot{q}_i(t)\right] \nonumber \\
& = & \sum^{n}_{i=1}m_i\frac{d}{dt}\left[\q_i(t)\,\times~\dot{q}_i(t)\right] \nonumber \\
& = & \sum^{n}_{i=1}\left[m_{i}\dot{\q}_i(t)\,\times~\dot{\q}_i(t) + \q_i(t)\,\times~m_{i}\ddot{\q}_i(t)\right] \nonumber \\
& = & \sum^{n}_{i=1}\left[\q_i(t)\,\times~F_i(t)\right]~\left(\textrm{since:}\quad ~a\times a\,\equiv\,0\right) \nonumber \\
& = & \sum^{n}_{i=1}\left[\q_i(t)\,\times~\sum_{j\,\neq\,i}F_{ij}(t)\right] \nonumber \\
& = & \sum^{n}_{i=1}\left[\q_i(t)\,\times~\sum_{j\,\neq\,i}f_{ij}(\mid\q_i(t)-\q_j(t)\mid)\frac{\q_i(t)-\q_j(t)}{\mid\q_i(t)-\q_j(t)\mid}\right] \nonumber \\
& = & \sum^{n}_{i=1}\left[\sum_{j\,\neq\,i}f_{ij}(\mid\q_i(t)-\q_j(t)\mid)\frac{\q_i(t)\,\times~\q_i(t)-\q_i(t)\,\times~\q_j(t)}{\mid\q_i(t)-\q_j(t)\mid}\right] \nonumber \\
& = & -\sum^{n}_{i=1}\sum_{j\,\neq\,i}\left[f_{ij}(\mid\q_i(t)-\q_j(t)\mid)\frac{\q_i(t)\,\times~\q_j(t)}{\mid\q_i(t)-\q_j(t)\mid}\right].
\label{6}
\end{eqnarray}
\end{flushleft}
But because $f_{ij}=f_{ji}$ and the quantity $\left[\q_i(t)\times\,\q_j(t)\right]$ is anti-symmetric, i.e. $\left[\q_i(t)\times\,\q_j(t)\right]=-\left[\q_j(t)\times\,\q_i(t)\right]$, the sum in \eqref{6} vanishes identically. Thus we have shown that,
\begin{center}
\fbox{
$\displaystyle\frac{d\J(t)}{dt}~=~-\sum^{n}_{i=1}\sum_{j\,\neq\,i}\left[f_{ij}(\mid\q_i(t)-\q_j(t)\mid)\frac{\q_i(t)\,\times~\q_j(t)}{\mid\q_i(t)-\q_j(t)\mid}\right]~=~0,$}
\end{center}
as was desired.
\end{document}
