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	\begin{document}

       \begin{center}
      {\large Classical Mechanics Homework\\}  
      {\small January 29, 2008 \\}  
      {\small John Baez\\}  
      {\small homework by: Scot Childress}  
      \end{center}

\section*{Conservation Laws for the $n$-Body Problem}



\newcommand{\eff}{\text{eff}}



\noindent 1.  Show that in an $n$-body system where the force of the $j$-th body on the $i$-th body is given by

\[F_{ij}(t) = f_{ij}(|q_i(t) - q_j(t)|)\lambda_{ij}(t)\]
(where $\lambda_{ij}$ denotes the unit vector in the direction $q_i - q_j$ at the time of interest and $f_{ij} = f_{ji}$; i.e. -- Newton's 3rd Law) the energy is conserved.



\vskip1em


Note that	$E(t) = T(t) + V(t).$  Let's work on $T$ first.  We're going to play fast and loose with notation; hopefully all computations will be clear.  Recall that $T$ is given as $T = \sum_i (1/2) m_i \dot q ^2$ so that


	\begin{align*}
			{dT\over dt} &= \sum_i m_i \dot q_i\cdot \ddot q_i \\
										&= \sum_i \dot q_i\cdot F_i & (\text{Newton's 2nd Law})\\
										&= \sum_i \sum_{i\neq j} \dot q_i \cdot F_{ij}\\
										&= \sum_i \sum_{i\neq j}  f_{ij}\cdot (\dot q_i\cdot \lambda_{ij})\\
										&= \sum_i \sum_{i <j} f_{ij} [\dot q_i\cdot \lambda_{ij} + \dot q_j\cdot \lambda_{ji}] &\text{(Newton's 3rd Law)}\\
										&= \sum_i \sum_{i<j} f_{ij} \cdot(\dot q_i - \dot q_j)\cdot \lambda_{ij}.
	\end{align*}
Now working with $V$ we have that:

	
	\begin{align*}
		{dV\over dt} &= \sum_i {dV_i\over dt}\\
			&= \sum_i \sum_{i < j} {dV_{ij}\over dt}\\
			&= -\sum_i\sum_{i < j} f_{ij} \cdot(\dot q_i - \dot q_j)\cdot \lambda_{ij}
		\end{align*}	
since $V_{ij}' = -f_{ij}.$  Summing the results of these two computations yields $E'(t) \equiv 0$, and hence energy is conserved.


\newpage

\noindent 2.  Show that angular momentum is conserved in the $n$-body problem.


	\begin{align*}
			J'(t)	&= \sum_i J_i'(t)\\
						&= \sum_i \dot q_i\times p_i + q_i\times \dot p\\
						&= \sum_i m_i(\dot q_i\times \dot q_i) + q_i\times F_i\\ 
						&= \sum_i \sum_{i\neq j} f_{ij}\cdot(q_i\times \lambda_{ij})\\
						&= \sum_i \sum_{i < j} f_{ij}\cdot (q_i \times \lambda_{ij} + q_j\times \lambda_{ji})\\
						&= \sum_i \sum_{i<j} {f_{ij}\over |q_i-q_j|} \cdot(q_i\times q_j + q_j\times q_i)
	\end{align*}
and the last term vanishes since $q_i\times q_j = -q_j\times q_i.$  (The second to last equality follows from Newton's 3rd Law.)
\end{document}










