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\begin{document}
\begin{center}
{\large Classical Mechanics Homework\\}
{\small February 7, 2008 \\}
{\small John Baez}
\end{center}
\vspace{1cm}
\noindent
{\it This homework gives you some choice in
which problems you do, depending on how ambitious you feel.
In the section on the Euclidean group you can either do
problems 1--3, or if you're feeling more ambitious, just problem 4.
In the section on the Galilei group you can either do problems
5--7, or if you're feeling more ambitious, just problem 8.
Everyone has to do problems 9--10.}
\vspace{1cm}
\section*{The Euclidean Group}
Recall that the {\bf orthogonal group} $\O(n)$ is the group of
linear transformations $R \maps \R^n \to \R^n$ that preserve
distances: $\|Rx\| = \|x\|$ for all $x \in \R$. In other words,
$\O(n)$ is the group of $n \times n$ matrices with $RR^* = 1$.
Define an element of the {\bf Euclidean group} $\E(n)$ to be a pair
$(R,u)$, where $R \in \O(n)$ and $u \in \R^n$. Any element $(R,u)$ gives
a transformation of $n$-dimensional Euclidean space
built from an orthogonal transformation and a translation:
\[ f_{(R,u)} \maps \R^n \to \R^n \]
defined by
\[ f_{(R,u)} (x) = Rx + u .\]
The map $f_{(R,u)}$ uniquely determines $R$ and $u$,
so we can also think of $\E(n)$ as a set of maps.
\vskip 1em
1. Given two elements $(R,u), (R',u') \in \E(n)$ show that
\[ f_{(R,u)} \circ f_{(R',u')} = f_{(R'',u'')} \]
for some unique $(R'',u'') \in \E(n)$. Work out the explicit
formula for $(R'',u'')$.
\vskip 1em
\noindent
{\it
This formula lets us define a `multiplication' operation
on $\E(n)$ by:
$(R,u) (R',u') = (R'', u'')$.
}
\vskip 1em
2. Given an element $(R,u) \in \E(n)$ show that
\[ f_{(R,u)}^{-1} = f_{(R',u')} \]
for some unique $(R',u') \in \E(n)$. Work out the explicit
formula for $(R',u')$.
\vskip 1em
\noindent
{\it
This formula lets us define an `inverse' operation
on $\E(n)$ by:
$ (R,u)^{-1} = (R', u') $.
}
\vskip 1em
3. With multiplication and inverses defined as above,
show that $\E(n)$ is a group.
(Hint: the good way to do this requires almost no calculation.)
\vskip 1em
\noindent
{\it
Note that as a set we have $\E(n) = \O(n) \times \R^n$.
However, as a group $\E(n)$ is not the direct product
of the groups $\O(n)$ and $\R^n$, because the
formulas for multiplication and inverse are {\rm not} just
\[ (R,u) (R',u') = (RR', u+u') , \qquad \qquad
(R,u)^{-1} = (R^{-1},-u) .\]
Instead, the formulas involve the action of $\O(n)$ on $\R^n$, so
we say $\E(n)$ is a `semidirect' product of $\O(n)$ and
$\R^n$.}
\vskip 1em
4. Suppose that $f \maps \R^n \to \R^n$ is a map that preserves
distances:
\[ |f(x) - f(y)| = |x - y| \]
for all $x,y \in \R^n$. Show that
\[ f(x) = Rx + u \]
for some $(R,u) \in \E(n)$. Thus we can more elegantly define
the Euclidean group to be the group of all distance-preserving
transformations of Euclidean space!
\section*{The Galilei Group}
Define an element of the {\bf Galilei group} $\G(n+1)$ to be
an triple $(f,v,s)$ where $f \in \E(n)$, $v \in \R^n$ and $s \in \R$.
We call $f$ a {\bf Euclidean transformation}, $v$ a {\bf Galilei
boost} and $s$ a {\bf time translation}.
Any element $(f,v,s) \in \G(n+1)$ gives a transformation of
$(n+1)$-dimensional spacetime
\[ F_{(f,v,s)} \maps \R^{n+1} \to \R^{n+1} \]
defined by
\[ F_{(f,v,s)} (x,t) = (f(x) + vt, t + s) \]
for all $(x,t) \in \R^{n+1}$.
The map $F_{(f,v,s)}$ uniquely determines $f,v$ and $s$,
so we can also think of $\G(n+1)$ as a set of maps.
\vskip 1em
5. Given two elements $(f,v,s), (f',v',s') \in \G(n+1)$ show that
\[ F_{(f,v,s)} \circ F_{(f',v',s')} = F_{(f'',v'',s'')} \]
for some unique $(f'',v'',s'') \in \G(n+1)$. Work out the explicit
formula for $(f'',v'',s'')$.
\vskip 1em
\noindent
{\it
This formula lets us define a `multiplication' operation
on $\G(n+1)$ by:}
$ (f,v,s) (f',v',s') = (f'', v'',s'') $.
\vskip 1em
6. Given an element $(f,v,s) \in \G(n+1)$ show that
\[ F_{(f,v,s)}^{-1} = F_{(f',v',s')} \]
for some unique $(f',v',s') \in \G(n+1)$. Work out the explicit
formula for $(f',v',s')$.
\vskip 1em
\noindent
{\it This formula lets us define an `inverse' operation
on $\G(n+1)$ by:}
$ (f,v,s)^{-1} = (f', v', s') $ .
\vskip 1em
7. With multiplication and inverses
defined as above, show that $\G(n+1)$ is a group.
(Again, the good way to do this requires almost no calculation.)
\vskip 1em
{\it
As a set we have $\G(n+1) = \E(n) \times \R^n \times \R$.
However, it is again not the direct product of these groups, but
only a semidirect product.}
\vskip 1em
8. Describe some structure on $\R^{n+1}$ such that $\G(n+1)$
is precisely the group of all maps $F \maps \R^{n+1} \to \R^{n+1}$
that preserve this structure. Prove that this is indeed the case.
\vskip 1em
\noindent
{\it More generally, we could axiomatically define an
$(n+1)$-dimensional {\bf Galilean spacetime} and prove that
the symmetry group of any such thing is isomorphic to $\G(n+1)$.
}
\section*{The Free Particle}
Recall that a group $G$ {\bf acts} on a set $X$ if for any
$g \in G$ and $x \in X$ we get an element $gx \in X$, and
\begin{equation}
g(g'x) = (gg')(x) , \qquad 1x = x
\label{action}
\end{equation}
for all $g,g' \in G$ and $x \in X$. We have just described
how the Euclidean group acts on Euclidean space and how the
Galilei group acts on Galilean spacetime. Now we will figure
out how the Galilei group acts on the phase space of a free particle!
Recall that the phase space of a particle in $n$-dimensional
Euclidean space is $X = \R^n \times \R^n$, where a point $(q,p) \in X$
describes the particle's position and momentum. I will tell
you how various subgroups of the Galilei group act on $X$, and
you will use that information to figure out how the the whole group
acts on $X$.
\vskip 1em
$\bullet$
The translation group $\R^n$ is a subgroup of $\E(n)$ and thus
$\G(n+1)$ in an obvious way, and it acts on $X$ as follows:
\[ u(q,p) = (q+u,p) \qquad \qquad u \in \R^n .\]
In other words, to translate a particle
we translate its position but leave its
momentum alone!
\vskip 1em
$\bullet$
The orthogonal group $\O(n)$ is also a subgroup of
$\E(n)$ and thus $\G(n+1)$ in an obvious way, and it acts on
$X$ as follows:
\[ R(q,p) = (Rq,Rp) \qquad \qquad R \in \O(n) .\]
In other words, to rotate a particle we rotate both its
position and momentum!
\vskip 1em
$\bullet$
The group of Galilei boosts $\R^n$ is a subgroup of $\G(n+1)$
in an obvious way, and it acts on $X$ as follows:
\[ v(q,p) = (q,p+mv) \qquad\qquad v \in \R^n .\]
In other words, to boost a particle's velocity by $v$ we
add $mv$ to its momentum but leave its position alone!
\vskip 1em
$\bullet$
Finally, the time translation group $\R$ is a subgroup
of $\G(n+1)$ in an obvious way, and it acts on $X$ as follows:
\[ s(q,p) = (q + sp/m, p) \qquad \qquad s \in \R .\]
This is where we are assuming the particle is {\bf free}:
the force on it is zero, so it moves along at a constant velocity,
namely $p/m$.
\vskip 1em
9. Assuming that all these group actions fit together to define an
action of the whole Galilei group on $X$, figure out how the whole
Galilei group acts on $X$.
\vskip 1em
\noindent
Hint: you'll probably want to use formula (\ref{action}) and also some
results from problems 1--3 and 6--7. An element of the Galilei
group is a triple $(f,v,s) \in \E(n) \times \R^n \times \R$, but here
it's best to think of it as a quadruple $(R,u,v,s) \in \O(n) \times \R^n
\times \R^n \times \R$, using the fact that $f = (R,u)$. I want you to
give me a formula like
\[ (R,u,v,s)(q,p) = \cdots \]
\vskip 1em
10. Finally, check that you {\it really have defined an action}
of $\G(n+1)$ on $X$. That is, check equation (\ref{action}) for
all $g = (R,u,v,s)$ and $g' = (R',u',v',s')$ in the Galilei group
and all $x = (q,p) \in X$.
\vfill
\end{document}