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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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