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%the Winter 2008 classical mechanics course by John Baez

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\newcommand{\p}{p}            %momentum
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\newcommand{\T}{T}            %kinetic energy
\newcommand{\V}{V}            %potential energy
\newcommand{\J}{J}            %angular momentum

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	\begin{document}

       \begin{center}
      {\large Classical Mechanics Homework\\}  
      {\small January 29, 2008 \\}  
      {\small John Baez}  
      \end{center}

\section{The 2-Body Problem}

The goal of this problem is to understand a pair of particles
interacting via a central force such as gravity.  We'll reduce it
to problem you've already studied --- the case of a {\it single}
particle in a central force.

Suppose we have a system of two particles interacting by a central force.  
Their positions are functions of time, say $\q_1,\q_2 \maps \R \to \R^3$,
satisfying Newton's law:
\[  m_1\ddot \q_1 = f(|\q_1 - \q_2|) {\q_1 - \q_2\over |\q_1 - \q_2|}  \]
\[  m_2 \ddot \q_2 = f(|\q_2 - \q_1|) {\q_2 - \q_1\over |\q_2 - \q_1|} . \]
Here $m_1,m_2$ are their masses, and the force is described by 
some smooth function $f \maps (0,\infty) \to \R$.   Let's write
the force in terms of a potential as follows:
\[       f(r) = -{d V\over dr} .\]

Using conservation of momentum and symmetry under translations
and Galilei boosts we can work in coordinates where
\be          m_1 \q_1(t) + m_2 \q_2(t) = 0 \label{com} \ee
for all times $t$.   This coordinate system is called the 
{\bf center-of-mass frame}.

We could use equation (\ref{com}) to solve for $\q_2$ in terms of
$\q_1$, or vice versa, but we can also use it to express both $\q_1$
and $\q_2$ in terms of the {\bf relative position}
\[                 \q(t) = \q_1(t) - \q_2(t) . \]
This is more symmetrical, so this is what we will do.   Henceforth we
only need to talk about $\q$.  Thus we have reduced the problem to a 
1-body problem!

\vskip2em
\noindent
{\textit{\textbf{Now here's where you come in:}}}

\vskip2em
1.   Show that $\q(t)$ satisfies the equation 
\[       m\ddot \q = f(|\q|) {\q\over |\q|}  \]
where $m$ is the so-called {\bf reduced mass}
\[              m = {m_1 m_2 \over m_1 + m_2}   .\]
Note that this looks exactly like Newton's second law
for a single particle!

\vskip1em

2.   Recall that the total energy $E$ of the 2-particle system is 
the sum of the kinetic energies of the particles plus the potential
energy.  Express $E$ in terms of $\q$ and the reduced mass.  
Show that 
\[     E =  {1\over 2}m \left|\dot \q\right|^2 + V(|\q|) \]
Note that this looks exactly like the energy of a single
particle!
\vskip1em

3.   Let $\J$ be the total angular momentum of the 2-particle system.
Show that 
\[   \J  = m \q \times \dot \q \]
Note that this looks exactly like the angular momentum
of a single particle!
\vskip1em

At this point we're back to a problem you've already solved:
a {\em single} particle in a central force.  The only difference
is that now $\q$ stands for the {\em relative} position and 
$m$ stands for the {\em reduced} mass!

\vskip 1em

So, we instantly conclude that two bodies orbiting each other 
due to the force of gravity will {\em both} have an orbit that's
either an ellipse, or a parabola, or a hyperbola... when viewed
in the center-of-mass frame.

\section{Poisson brackets}

Let $\R^{2n}$ be the {\bf phase space} of a particle in 
$\R^n$, with coordinates $q_i, p_i$ ($1 \le i \le n$).
Let $C^\infty(\R^{2n})$ be the set of smooth real-valued
functions on $\R^{2n}$, which becomes an commutative algebra
using pointwise addition and multiplication of functions.  

\vskip 1em

We define the {\bf Poisson bracket} of
functions $F, G \in C^\infty(\mathbb{R}^{2n})$ by:
\[        \{ F, G \} = \sum_{i = 1}^n 
\frac{\partial F}{\partial p_i} \frac{\partial G}{\partial q_i} - 
\frac{\partial G}{\partial p_i} \frac{\partial F}{\partial q_i}
.\]

\vskip 1em
4.  Show that Poisson brackets make the vector space
$C^\infty(R^{2n})$ into a {\bf Lie algebra}.  In other 
words, check the {\bf antisymmetry} of the bracket:
\[        \{F,G\} = -\{G,F\}  \]
the {\bf bilinearity} of the bracket:
\[        \{F, \alpha G + \beta H \} = 
          \alpha \{F,G\} + \beta \{F,H\}  \]
\[        \{\alpha F + \beta G, H \} = 
          \alpha \{F,H\} + \beta \{G,H\}  \]
and {\bf Jacobi identity}:
\[     \{F, \{G, H \} \} = 
      \{\{F,G\}, H\} + \{G, \{F, H \} \}  \]
for all $F,G,H \in C^\infty(\R^{2n})$ and $\alpha, \beta \in \R$.

\vskip 1em

({\it Note the Jacobi identity resembles the product
rule $d(GH) = (dG)H + GdH$, with bracketing by $F$ playing
the role of $d$.   This is no accident!})

\vskip 1em
5.  Show that Poisson brackets and ordinary 
multiplication of functions make the vector space $C^\infty(\R^{2n})$ 
into a {\bf Poisson algebra}.  This is a Lie algebra that is 
also a commutative algebra, with the bracket $\{F,G\}$ and
the product $FG$ related by the {\bf Leibniz identity}:
\[       \{ F, GH \} = \{F,G\} H + G \{F,H\}  .\]
\vskip 1em
({\it Again this identity resembles the product rule!})

\end{document}

