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\begin{document}
\begin{center}
{\large Classical Mechanics Homework\\}
{\small February 2, 2008 \\}
{\small John Baez\\}
{\small Homework by: Scot Childress}
\end{center}
\section*{The $2$-Body Problem}
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!
\vskip1em
\noindent 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}
\[ m = {m_1m_2\over m_1 + m_2}.\]
Using the expressions for $m_i \ddot q_i$ in terms of $f$ given in the introduction, we have:
\begin{align*}
m\ddot q &= {m_1 m_2\over m_1 + m_2} (\ddot q_1 - \ddot q_2)\\
&= {1\over m_1 + m_2}\left(m_2 f(|q|){q\over |q|} + m_1 f(|q|){q\over |q|}\right)\\
&= f(|q|){q\over |q|}.
\end{align*}
\vskip1em
\noindent 2. Show that
\[E = {1 \over 2} m |\dot q|^2 + V(|q|).\]
Recall that the the kinetic energy of the $i$-th particle is given by
\[T_i = {1\over 2} m \dot q_i^2\]
and that the total kinetic energy for the system is $T = T_1 + T_2.$ Bearing this in mind, we calculate:
\begin{align}
{1\over 2} m\dot q^2 &= {1\over 2}{m_1m_2\over m_1 + m_2} (\dot q_1 - \dot q_2)^2\nonumber\\
&={1\over m_1 + m_2}(m_2T_1 + m_1T_2 - m_1m_2\dot q_1 \dot q_2).\label{kinetic1}
\end{align}
Note that \eqref{com} implies that
\be\label{switcheroo}\dot q_i = -{m_j\over m_i}\dot q_j\ee
for $q_i$ equal to $q_1$ or $q_2$. Hence
\[{1\over 2}m_1m_2\dot q_1 \dot q_2 = -{1\over 2} m_i^2\dot q_i^2 = -m_i T_i\]
for $i = 1,2.$ Thus \eqref{kinetic1} becomes
\[{1\over m_1 + m_2}(m_2T_1 + m_1T_2 - m_1m_2\dot q_1 \dot q_2) = {1\over m_1 + m_2}\left(m_2T_1 + m_1T_2 + (m_1 T_1 + m_2 T_2)\right) = T_1 + T_2.\]
Whence
\[{1\over 2} m\dot q^2 = T_1 + T_2 = T,\]
and the energy of the system is given by
\[E = T + V(|q_1 - q_2|) = {1\over 2} m\dot q^2 + V(|q|).\]
\vskip1em
\noindent 3. Establish
\[J = m q\times \dot q\]
where $J$ is the total angular momentum.
We work from the right hand side:
\begin{align*}
m q\times \dot q &= {m_1m_2\over m_1 + m_2} (q_1 - q_2)\times (\dot q_1 - \dot q_2)\\
&= {m_1m_2\over m_1 + m_2}(q_1\times \dot q_1 - q_1\times\dot q_2 - q_2\times \dot q_1 + q_2\times \dot q_2)\\
&= {m_1m_2\over m_1 + m_2} \left(q_1\times \dot q_1 + {m_1\over m_2}q_1\times\dot q_1 + {m_2\over m_1} q_2\times \dot q_2 + q_2\times \dot q_2\right)\\
&= {m_1m_2\over m_1 + m_2}\left({m_1 + m_2\over m_2} q_1\times \dot q_1 + {m_1 + m_2\over m_1} q_2\times \dot q_2\right)\\
&= m_1 q_1\times \dot q_1 + m_2 q_2\times \dot q_2\\
&= J_1 + J_2\\
&= J;
\end{align*}
where the third equality follows from \eqref{switcheroo}.
\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
\noindent 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 $\alpha, \beta \in \R$.
\vskip1em
To make life a whole lot easier on ourselves we will impose the following conventions:
\begin{enumerate}
\item Lower indices indicate differentiation with respect to $p_i$, for instance:
\[F_i = {\partial F \over \partial p_i}.\]
\item Upper indices indicate differentiation with respect to $q_i$, that is:
\[G^i = {\partial G \over \partial q_i}.\]
\item All indexed quantities are assumed to represent sums. For instance:
\[ F^i_j G_i = \sum_{i = 1}^n \sum_{j = 1}^n {\partial F \over \partial p_j \partial q_i} {\partial G\over \partial i}.\]
\end{enumerate}
Now then, let's get down to business. First, we will show that $\{\cdot, \cdot\}$ anti-commutes. This is simple, since
\[\{F, G\} = F_iG^i - G_i F^i = -(G_iF^i - F_i G^i) = -\{G, F\}.\]
Bilinearity is likewise trivial:
\[\{F, aG + bH\} = F_i(aG + bH)^i - (aG + bH)_i F^i = a(F_iG^i - G_iF^i) + b(F_iH^i - H_i F^i) = a\{F, G\} + b\{F, H\}\]
(\emph{and similarly for the right hand slot of} $\{\cdot,\cdot\}$).
Now for the rough part of establishing the Jacobi identity. We will see that our conventions will reduce the problem to one of accounting---surely there is a more elegant way to establish the identity, but sometimes it is nice to get our hands dirty with actual computation. We will write down only the significant portions of the computation, leaving the reader to fill in the gaps. Here we go:
\begin{align*}
\{\{F,G\}, H\} + \{G, \{F, H\}\} &= \{F, G\}_i H^i - H_i\{F,G\}^i + G_i\{F, H\}^i - \{F, H\}_iG^i\\
&= (F_jG^j - G_jF^j)_i H^i - H_i(F_jG^j - G_j F^j)^i + G_i(F_j H^j - H_j F^j)^i\\
& \hspace{2.75 in} - (F_j H^j - H_jF^j)_iG^i.
\end{align*}
At this stage, we complete the differentiation and collect first and second partials of $F$ (using the fact that $F$ is smooth---i.e.- the mixed second partials of $F$ are equal) to obtain:
\[ F_{ij}(G^jH^i - G^iH^j) + F^i_j[(G_iH^j - G_jH^i) + (G^iH_j - G^jH_i)] + F^{ij}(H_i G_j - H_jG_i)\]
\[\hspace{.5 in}+ F_j(G^j_iH^i -H_iG^{ij} + G_i H^{ij} - G^i H^j_i) + F^j(G^i_j H_i - G_iH^i_j + G^iH_{ij} - G_{ij}H^i). \]
Now, the first three summands of the above vanish. To see this, note that if $i=j$ then each term is $0$, and if $i\neq j$ exchanging $i$ with $j$ makes the terms negative: hence, in the sum, each of these three terms come in opposite pairs. Taking this cancellation into account, and noting that the last two terms of the sum are actually
\[F_j(G_iH^i - G^iH_i)^j - (G_iH^i - G^iH_i)_jF^j= F_j\{G, H\}^j - \{G, H\}_j F^j,\]
we see that
\[ \{\{F,G\}, H\} + \{G, \{F, H\}\} = \{F, \{G, H\}\}.\]
\vskip1em
\noindent 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}:
\[ \{ F, GH \} = \{F,G\} H + G \{F,H\} .\]
Again, the conventions we established above will set us free:
\[\{ F, GH \} = F_i(GH)^i - (GH)_i F^i = F_i(G^iH + GH^i) - (G_iH + GH_i)F^i\]
\[\hspace{2 in} = (F_iG^i - G_iF^i)H + G(F_iH^i - H_iF^i) = \{F, G\}H + G\{F, H\}.\]
\end{document}