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\begin{document}
\begin{center}
{\large Classical Mechanics Homework\\}
{\small February 5, 2$\infty$8 \\}
{\small John Baez}
{\small homework by C.Pro}
\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
\vskip2em
1. Now by taking the second time derivative of the relative postion and using
equation (\ref{com}) to solve for $\dot\q_1$ in terms of $\dot\q_2$, we have
\ban
\ddot\q&=&\ddot\q_1-\ddot\q_2\\
&=&\left(1+{m_1\over m_2}\right)\ddot\q_1\\
&=&\left({m_1+m_2\over m_1m_2}\right)m_1\ddot\q_1\\
&=&\left({m_1+m_2\over m_1m_2}\right)
f(|\q_1 - \q_2|) {\q_1 - \q_2\over |\q_1 - \q_2|}
\ean
where the last equality follows from Newton's 2$^{\mathrm{nd}}$ law. But note
that this says:
$$m\ddot \q = f(|\q|) {\q\over |\q|}$$
where $m$ is the so-called {\bf reduced mass}
$$ m = {m_1 m_2 \over m_1 + m_2}, $$
and this looks exactly like Newton's second law for a single particle!
\vskip1em
2. We also have a similar single particle property involving the total energy.
Indeed, by squaring both sides of equation (\ref{com})
then adding $m_1m_2(\dot\q_1^2-\dot\q_2^2)$ to both sides, we have
$$
m_1^2\dot\q_1^2+m_2^2\dot\q_2^2+2m_1m_2\dot\q_1\dot\q_2
+m_1m_2(\dot\q_1^2-\dot\q_2^2)
=m_1m_2(\dot\q_1^2-\dot\q_2^2),
$$
or equivalently,
$$m_1(m_1+m_2)\dot\q_1^2+m_2(m_1+m_2)\dot\q_2^2= m_1m_2(\dot\q_1-\dot\q_2)^2.$$
Now note the left side is $2(m_1+m_2)$ times the total kinetic energy, thus we have a new way
to write this total kinetic energy in terms of the reduced mass and relative postion, that is
$${1\over2}m_1\dot\q_1^2+{1\over2}m_2\dot\q_2^2={1\over 2}m|\dot\q|^2.$$
Therefore, the total energy can be written as
$$E = {1\over 2}m \left|\dot \q\right|^2 + V(|\q|)$$
and this looks exactly like the energy of a single particle!
\vskip1em
3. It should be no suprise that we'll obtain a similar suprise for the total
angular momentum $\J$ of these two bodies. Indeed,
\ban
\J &=&m_1\q_1\times\dot\q_1 + m_2\q_2\times\dot\q_2\\
&=&m\left(1+{m_1\over m_2}\right)q_1\times\dot\q_1 + m\left(1+{m_2\over m_1}\right)q_1\times\dot\q_2\\
&=&m\q_1\times\dot\q_1+\q_1\times{m_1\over m_2}\dot\q_1+
m\q_2\times\dot\q_2+\q_2\times{m_2\over m_1}\dot\q_2\\
&=&m\q_1\times(\dot\q_1-\dot\q_2)-m\q_2\times(\dot\q_1-\dot\q_2)\\
&=&m\q\times\dot\q
\ean
and this looks exactly like the angular momentum of a single particle!
\vskip1em
At this point we're back to a problem we'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.
\newpage
\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. The Poisson brackets make the vector space $C^\infty(R^{2n})$ into a {\bf Lie algebra}.\\
\textit {Proof: } Antisymmetry of the bracket is immediate from the definition and bilinearity
follows from the linearity of the differential operator. However, some work is required to show
the Jacobi identity. We first assume that in which we are asked to prove in question 5,
namely that the Poisson bracket and the multiplication in the commutative algebra
$C^{\infty}(\R^{2n})$ are related through the Leibniz identity. Let $A, B, C\in C^{\infty}(\R^{2n})$,
for completeness we note:
$${\partial \over \partial x}\{A,B\} =
\sum_{i=1}^n
\left(
{\partial^2A \over \partial x\partial\p_i}{\partial B \over \partial\q_i}+
{\partial A \over \partial\q_i}{\partial^2B \over \partial x\partial\p_i}
\right) -
\left(
{\partial^2A \over \partial x\partial\q_i}{\partial B \over \partial\p_i} +
{\partial A \over \partial\p_i}{\partial^2B \over \partial x\partial\q_i}
\right)
$$
and
$$\{\{A,B\},C\}=
\sum_{i=1}^n
{\partial \{A,B\}\over \partial \p_i} {\partial C \over \partial \q_i} -
{\partial \{A,B\}\over \partial \q_i}{\partial C\over \partial \q_i}.
$$
We now see that a permutation in the letters $G$ and $H$ in the expression $\{\{F,G\},H\}$
fixes all terms that contain a factor of $F$ under a second order differential operator. Therefore,
in the expression $\{\{F,G\},H\}-\{\{F,H\},G\}$, these terms are killed and thus can be written in
the following form:
$$
\sum_{i=1}^n
\left(
\left\{F,{\partial G\over \partial p_i }\right\}{\partial H\over \partial q_i}-
\left\{F,{\partial G\over \partial q_i }\right\}{\partial H\over \partial p_i}
\right)-
\left(
\left\{F,{\partial H\over \partial p_i }\right\}{\partial G\over \partial q_i}-
\left\{F,{\partial H\over \partial q_i }\right\}{\partial G\over \partial p_i}
\right).
$$
However, the Leibinz identity says this can be reduced further to
$$
\sum_{i=1}^n
\left(
\left\{F,{\partial G\over \partial p_i }{\partial H\over \partial q_i}\right\}-
\left\{F,{\partial G\over \partial q_i }{\partial H\over \partial p_i}\right\}
\right).
$$
Now anitsymmetry and bilinearity give us the following:
\ban
\{\{F,G\},H\}+\{G,\{F,H\}\}&=&\{\{F,G\},H\}-\{\{F,H\},G\}\\
&=& \left\{F,\sum_{i=1}^n
{\partial G\over \partial p_i }{\partial H\over \partial q_i}-
{\partial G\over \partial q_i }{\partial H\over \partial p_i}\right\}\\
&=&\{F,\{G,H\}\}
\ean
and this is the Jacobi identity. To complete the proof we are left to independently verify the
Leibniz identity. But since no one is actually reading this, it will suffice to sit in my chair and
shout out loud the word refrigerator as I type it.
({\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. The Poisson brackets and ordinary
multiplication of functions make the vector space $C^\infty(\R^{2n})$
into a {\bf Poisson algebra}.\\
\textit{Proof: } We need only to verify the Leibniz identity:
\[ \{ F, GH \} = \{F,G\} H + G \{F,H\} .\]
Indeed,
\ban
\{ F, GH \}&=&\sum_{i=1}^n
{\partial F\over \partial p_i }{\partial GH\over \partial q_i}-
{\partial F\over \partial q_i }{\partial GH\over \partial p_i}\\
&=&\sum_{i=1}^n
{\partial F\over \partial p_i }
\left({\partial G\over \partial q_i}H+G{\partial H\over \partial q_i}\right)-
{\partial F\over \partial q_i }
\left({\partial G\over \partial p_i}H+G{\partial H\over \partial p_i}\right)\\
&=&\left(\sum_{i=1}^n
{\partial F\over \partial p_i }{\partial G\over \partial q_i}-
{\partial F\over \partial q_i }{\partial G\over \partial p_i}\right) H +
G\left(\sum_{i=1}^n
{\partial F\over \partial p_i }{\partial H\over \partial q_i}-
{\partial F\over \partial q_i }{\partial H\over \partial p_i}\right)\\
&=&\{F,G\} H + G \{F,H\}.
\ean
\vskip 1em
({\it Again this identity resembles the product rule!})
\end{document}