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\begin{document}
\title{The Asymmetric Top}
\maketitle
We will ``approximate'' the asymmetric top by a finite set of point
masses which are attached to each other by massless, rigid rods
which ensure that the motion of the whole ensemble is described by a
path in the group of Euclidean motions in space. The exposition given
below is similar to that which can be found in the book on Classical
Mechanics by V.~I.~Arnold.
For the $i$-th point mass with mass $m_i$, its position $\vec{r}_i$ is a
function of time $t$ given by the formula
\[ \vec{r}_i(t)=g(t)\cdot \vec{v}_i + \vec{r}(t) \]
where $\vec{v}_i$ is a constant vector and $g$ is a path in the
orthogonal group. Let $M=\sum_i m_i$ denote the total mass. The
position $\vec{R}$ of the centre of mass is given by
\[ \vec{R}=\sum_i m_i \vec{r}_i/M. \]
Moreover, if $\vec{V}=\sum_i m_i \vec{v}_i/M$, then we have
\[ \vec{R}(t)=g(t)\cdot \vec{V} + \vec{r}(t). \]
If we set $\vec{u}_i=\vec{v}_i-\vec{V}$, then we obtain
\[
\begin{split}
\vec{r}_i(t) & = g(t)\cdot\vec{u}_i + \vec{R}(t) \\
& = \text{~motion \emph{about} the centre of mass~} +
\text{~motion \emph{of} the centre of mass}
\end{split}
\]
We note that $\sum_i m_i \vec{u}_i = \vec{0}$.
Now let us consider the velocity vectors of individual point masses,
\[ \dot{\vec{r}}_i(t) = \dot{g}(t)\cdot \vec{u}_i + \dot{\vec{R}}(t) \]
Since $g(t)$ is a path in the space of orthogonal matrices, we see that
$g(t)^{-1}\dot{g}(t)$ is a skew-symmetric matrix. Hence there is a vector valued
function $\vec{\Omega}(t)$ so that for any vector $\vec{u}$ we have
\[ \vec{\Omega}(t) \times \vec{u} = g(t)^{-1}\cdot\dot{g}(t)\cdot \vec{u} \]
$\vec{\Omega}(t)$ is called the \emph{angular velocity} about the centre of mass in the
\emph{body frame}. If $\vec{\omega}=g(t)\cdot\vec{\Omega}$, then by applying $g(t)$
on all terms of the above equation, we see that for any vector $\vec{r}$ we have
\[ \vec{\omega}(t) \times \vec{r} = \dot{g}(t)\cdot g(t)^{-1}\cdot\vec{r} \]
$\vec{\omega}(t)$ is called the angular velocity about the centre of mass in the
{\em stationary frame}. Thus, we can re-write the velocity vector of the $i$-th
point mass
\[ \dot{\vec{r}}_i(t) = \vec{\omega}(t)\times (\vec{r}_i-\vec{R}(t))
+ \dot{\vec{R}}(t) \]
The momentum of the system as a whole is given by
\[ \vec{p} = \sum_i m_i \dot{\vec{r}}_i = M \dot{\vec{R}} \]
The angular momentum of the system as a whole in the {\em stationary} co-ordinates
is given by
\begin{multline*}
\vec{\lambda}_{\rm tot} = \sum_i m_i \vec{r}_i\times\dot{\vec{r}}_i
= \sum_i m_i (\vec{r}_i-\vec{R})\times\dot{\vec{r}}_i +
\sum_i m_i \vec{R}\times\dot{\vec{r}}_i\\
= \sum_i m_i (\vec{r}_i-\vec{R})\times(\vec{\omega}\times (\vec{r}_i-\vec{R}(t))) +
M \vec{R}\times\dot{\vec{R}}
\end{multline*}
(the remaining terms vanish because $\sum_i m_i \vec{r}_i = M
\vec{R}$). The latter term in the above expression for
$\vec{\lambda}_{\rm tot}$ has the obvious interpretation as the
angular momentum of (a point mass concentrated at) the centre of
mass. The former term
\[
\vec{\lambda} = \sum_i m_i (\vec{r}_i-\vec{R})
\times (\vec{\omega}(t)\times (\vec{r}_i-\vec{R}))
\]
is referred to as the angular momentum of the system \emph{about}
the centre of mass in the \emph{stationary frame}. Applying $g(t)^{-1}$ to the entire expression
gives
\[
\vec{\Lambda} = \sum_i m_i \vec{u}_i\times(\vec{\Omega}\times
\vec{u}_i)
\]
which is called the angular momentum in the \emph{body frame}. We
note that the map
\[
\vec{w} \mapsto I(\vec{w})=\sum_i m_i
\vec{u}_i\times(\vec{w}\times\vec{u}_i)
\]
depends only on the initial position of the point masses with respect
to their centre of mass and is thus associated to the configuration
or ``shape'' of this system; $I$ is called the \emph{moment of
inertia} or more strictly, the moment of inertia \emph{tensor} of the
configuration of the point masses.
We can also compute the total kinetic energy $T_{\rm tot}$ of the system
\begin{multline*}
T_{\rm tot} = \half\sum_i m_i ||\dot{\vec{r}}_i||^2 =
\half\sum_i m_i (\dot{\vec{r}}_i-\dot{\vec{R}}).\dot{\vec{r}}_i +
\half\sum_i m_i \dot{\vec{R}}.\dot{\vec{r}}_i \\
= \half\sum_i m_i
(\dot{\vec{r}}_i-\dot{\vec{R}}).(\dot{\vec{r}}_i-\dot{\vec{R}}) +
\half M \dot{\vec{R}}.\dot{\vec{R}} \\
= \half\sum _i m_i ||\vec{\omega}\times(\vec{r}_i-\vec{R})||^2 +
\half M ||\dot{\vec{R}}||^2
\end{multline*}
The latter term is the kinetic energy of a point mass concentrated at the
centre of mass. The former term
\[
T = \half\sum _i m_i ||\vec{\omega}\times(\vec{r}_i-\vec{R})||^2
= \half\sum _i m_i ||\vec{\Omega}\times\vec{u}_i||^2
\]
is called the kinetic energy \emph{about} the centre of mass or the
\emph{rotational} kinetic energy. Note that
\[ T = \half\sum _i m_i
\vec{\Omega}.(\vec{u}_i\times(\vec{\Omega}\times\vec{u}_i))
= \half\vec{\Omega}.I(\vec{\Omega})
\]
Using the identity
\[ \vec{x}.(\vec{y}\times\vec{z}) = \vec{y}.(\vec{z}\times\vec{x}) \]
we obtain $\vec{x}.I(\vec{w})=\vec{w}.I(\vec{x})$; in other words $I$
is given by a symmetric matrix. We also see that
\[
\vec{w}.I(\vec{w}) = \sum_i m_i ||(\vec{w}\times\vec{u}_i)||^2
= ||\vec{w}||^2\sum_i m_i
\text{~(distance of $\vec{u}_i$ from $\bbR\vec{w}$)}^2
\]
Thus for a non-zero vector $\vec{w}$, the latter quantity
\[ I_{\vec{w}} = \sum_i m_i
\text{~(distance of $\vec{u}_i$ from $\bbR\vec{w}$)}^2
\]
is sometimes called the moment of inertia in the direction $\vec{w}$;
it does not depend on the magnitude of $\vec{w}$. Because $I$ is
symmetric, there is an orthonormal eigen-basis
$\{\vec{e}_1,\vec{e}_2,\vec{e}_3\}$ for $I$. The numbers
$I_i=I_{\vec{e_i}}$ are called the principal moments of the system.
The behaviour of the $i$-th point mass is dependent on the force
$\vec{F}_i$ acting on it. Newton's law states that $\vec{F}_i = m_i
\ddot{\vec{r}}_i$. From the expression for $\dot{\vec{r}}_i$ and the
identity $\dot{\vec{\omega}}=g\cdot\dot{\vec{\Omega}}$ we obtain
\begin{multline*}
\ddot{\vec{r}}_i
= \dot{g}\cdot(\vec{\Omega}\times \vec{u}_i)
+ g\cdot(\dot{\vec{\Omega}}\times \vec{u}_i) + \ddot{\vec{R}} \\
= g\cdot(\vec{\Omega}\times(\vec{\Omega}\times \vec{u}_i))
+ g\cdot(\dot{\vec{\Omega}}\times \vec{u}_i) + \ddot{\vec{R}} \\
= \vec{\omega}\times(\vec{\omega}\times (\vec{r}_i-\vec{R}))
+ \dot{\vec{\omega}}\times (\vec{r}_i -\vec{R}) + \ddot{\vec{R}}
\end{multline*}
Let $\vec{F}=\sum_i \vec{F}_i$. Then we have $\vec{F}=M\ddot{\vec{R}}$ so
that the centre of mass of the system behaves as if the force $\vec{F}$ is
acting on (a point mass centred at) it. Now we compute the rate of
change of angular momentum,
\begin{multline*}
\dot{\vec{\lambda}}_{\rm tot} =
\sum_i m_i \vec{r}_i\times\ddot{\vec{r}}_i
= \sum_i m_i (\vec{r}_i -\vec{R})\times\ddot{\vec{r}}_i +
\sum_i m_i \vec{R}\times\ddot{\vec{r}}_i \\
= \sum_i (\vec{r}_i - \vec{R}) \times \vec{F}_i +
\vec{R} \times \vec{F}
\end{multline*}
The former term is called the \emph{torque} acting on the system
about the centre of mass in the stationary frame. If we split
$\vec{F}_i$ as $\vec{G}_i+\vec{H}_i$, where $\vec{G}_i$ is towards
the centre of mass (i.~e.\ parallel to the vector
$\vec{r}_i-\vec{R}$) and $\vec{H}_i$ is orthogonal to it; then
clearly the torque is given by $\sum_i
(\vec{r}_i-\vec{R})\times\vec{H}_i$. As before we can apply
$g(t)^{-1}$ to it to obtain $\vec{\tau}=\sum_i \vec{u}_i \times
\vec{h}_i$ as the torque acting on the top in the body frame; here
$\vec{h}_i=g^{-1}\vec{H}_i$. On the other hand, using the above
expression for $\ddot{\vec{r}}_i$ we see that
\begin{multline*}
\dot{\vec{\lambda}}=
\sum m_i (\vec{r}_i - \vec{R}) \times
(\ddot{\vec{r}}_i-\ddot{\vec{R}})
= \sum m_i (\vec{r}_i - \vec{R}) \times \ddot{\vec{r}}_i \\
= \sum m_i (\vec{r}_i - \vec{R})\times
\left(
\vec{\omega}\times(\vec{\omega}\times(\vec{r}_i-\vec{R}))
+ \dot{\vec{\omega}}\times(\vec{r}_i-\vec{R})
\right) \\
= \vec{\omega} \times \vec{\lambda} +
\sum m_i (\vec{r}_i - \vec{R})\times
(\dot{\vec{\omega}}\times(\vec{r}_i-\vec{R}))
\end{multline*}
We apply $g^{-1}$ to express everything in the body frame.
\[ \vec{\tau} = \vec{\Omega} \times \vec{\Lambda} +
\dot{\vec{\Lambda}}
\]
This expression in the body frame for the derivative of the angular
momentum in terms of the torque and the angular velocity is called
Euler's equation.
To summarise, the motion of an asymmetric top can separated into two
components, the motion of its centre of mass and the motion about the
centre of mass. The centre of mass behaves exactly like a point of mass
$M$ with position $\vec{R}$ subject to a force $\vec{F}$ under
Newton's equations. The motion
about the centre of mass is described in the body frame by a path $g$
is the group of rotations, the moment of inertia tensor $I$ and torque
$\tau$ to which the body is subjected. The equation of motion in this
case is Euler's equation. Since the motion of a point mass in a
Newtonian system is well understood, we will concentrate on the latter.
The only expression that depends on the distribution of masses is $I$.
Replacing the summation by integration and the masses by a mass density
$\rho$, we see that another expression for the moment of inertia is
\[
I(\vec{\Omega}) = \iiint \rho~
\vec{r}\times(\vec{\Omega}\times\vec{r})~dx\,dy\,dz
\]
Similarly, we see that
\[
I_{\vec{w}} = \iiint \rho~d(\vec{r},\bbR\vec{w})^2~dx\,dx\,dz
\]
where $d(\vec{r},\bbR\vec{w})$ denotes the distance of $\vec{r}$ from
the line $\bbR\vec{w}$. If we perform these computations for a
ellipsoid with principal axes being the co-ordinate axes with principal
lengths $a$, $b$ and $c$, then
\[
(x,y,z).I((x,y,z)) = \text{const.}\cdot (
(b^2+c^2)x^2
(c^2+a^2)y^2
(a^2+b^2)z^2
)
\]
In particular, note that the ``shape'' of the surface
$\vec{w}.I(\vec{w})=1$ is quite different from that of the boundary of
the given ellipsoid.
We say that the rotation is \emph{inertial} if the torque vanishes;
i.~e.\ $\vec{\tau}=0$.
It follows that $\dot{\vec{\lambda}}=0$ so that the angular
momentum $\vec{\lambda}$ in the stationary frame is conserved. In
particular, the magnitude $||\vec{\Lambda}||=||\vec{\lambda}||$ of the
angular momentum in the body frame is preserved. Moreover, taking the
derivative of the rotational kinetic energy
$T=\half\vec{\omega}.\vec{\lambda}=\half\vec{\Omega}.\vec{\Lambda}$
we see (using the symmetry of $I$) that
% \begin{multline*}
\[
\dot{\vec{\Omega}}.\vec{\Lambda}=\dot{\vec{\Omega}}.I(\vec{\Omega})=
(\half\vec{\Omega}.I(\dot{\vec{\Omega}})
+ \half\dot{\vec{\Omega}}.I(\vec{\Omega}))
=\dot{T}=\half\dot{\vec{\omega}}.\vec{\lambda} =
\half\dot{\vec{\Omega}}.\vec{\Lambda}
\]
% \end{multline*}
Thus $\dot{\vec{\Omega}}.\vec{\Lambda}=\dot{T}=0$ and the so kinetic
energy $T$ in the body frame is preserved. It follows that
$\vec{\Omega}$ lies on the intersection of the sphere $S$ defined by
$||\vec{\Omega}||=\text{const.}$ and the ellipsoid $E$
defined by $\vec{\Omega}.I(\vec{\Omega})=\text{const.}$; the ellipsoid
could be degenerate in case $I$ is not invertible (but this won't happen
for ``solid'' tops).
Poinsot offered a more precise description as follows. Consider the
image $g\cdot E$ of the ellipsoid. The vector
$\vec{\omega}=g\cdot\vec{\Omega}$ lies on this ellipsoid. Moreover, a
vector $\vec{w}$ is tangent to $E$ at $\vec{\Omega}$ if
$\vec{w}.\vec{\Lambda}=\vec{w}.I(\vec{\Omega})=0$. Thus a vector
$\vec{x}$ is tangent to $g\cdot E$ at $\vec{\omega}$ if
$\vec{x}.\vec{\lambda}=0$. Said differently, the tangent plane $\pi$ to
$g\cdot E$ at $\vec{\omega}$ consists of vectors $\vec{y}$ such that
$\vec{y}.\vec{\lambda}=\vec{\omega}.\vec{\lambda}$. We recognise the
latter expression as $2T$, twice the rotational kinetic energy, which
is a constant of motion. In other words, $\pi$ is also a constant of
motion.
To summarise, the rotational motion $g$ is such that the moving
ellipsoid $g\cdot E$ remains tangent to a fixed plane and the point of
tangency provides the axis of rotation; such a motion of the ellipsoid
$E$ is called rolling without slipping on the plane $\pi$. To recover
the rotational motion of the original top we note that
$(\vec{r}_i-\vec{R})=g\cdot u_i$; so the top is ``affixed'' to the
ellipsoid through its centre of mass with the body frame aligned so
that the eigen-basis of the moment of inertia are the principal axes of
the ellipsoid. {\em Warning}: the reader should beware that we are {\em
not} describing the inertial rotational motion of a top shaped like
$E$---rather the motion of $E$ described gives a nice geometric
description of the motion of the original top.
\end{document}
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