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\title{
A Geometric Framework for time-dependent Billiards}

\author
{Jair Koiller\thanks{On a leave to the Instituto de Matem\'aticas y 
F\'\i sica Fundamental, CSIC, Madrid, Espa\~na, at the final stages of
this work, under DGICYT grant SAB-0105 and travel financed by
IBM-Brasil} \\
Laborat\'orio Nacional de Computa\c c\~ao Cient\'\i fica\\
and 
Instituto de Matem\'atica, UFRJ, Brasil.
\and
Roberto Markarian\thanks{Partially supported by Fac. de Ciencias, Uruguay}
\\
Inst. de Matem\'atica y Estad\'\i stica 
''Prof.Ing. Rafael Laguardia'' \\
Fac. de Ingenier\'\i a, Uruguay.
\and
Sylvie Oliffson Kamphorst\\
S\^onia Pinto de Carvalho\\
Departamento de Matem\'atica, ICEx, UFMG, Brasil.}


\date{}
\begin{document}
\maketitle

\centerline{
{Keywords:} billiards, symplectic structures, linearization, stability.
}

\begin{abstract}
 Billiards with moving boundaries can be viewed as the limiting
 case of rigid ones, constructed on an augmented configuration space.
 The metric is defined by a new kinetic energy, singular in the limit.
 We describe the symplectic structure  of the time-dependent billiard map
 and specify suitable coordinates for the convex case of dimension 2.
\end{abstract}


\section{Introduction}
Billiard flows are defined by the free motion of a point mass in a
connected domain with piecewise smooth boundary on a 
 Riemannian manifold.
The motion is along the geodesics inside the domain and the particle is
 reflected elastically when it reaches the smooth parts of the 
boundary.
 {\em Billiards} aroused  mathematical interest  since
 G.D.Birkhoff used them to  illustrate many geometrical ideas
 for Dynamical  Systems  \cite{kn:bir}.
The studies of billiards gained further impulse  when Krylov and Sinai
related them with the ergodic hypothesis of Boltzman-Gibbs \cite{kn:kryl}.
 For surveys and a complete introduction to 
 the subject, with emphasis on
 ergodic properties, see \cite{kn:corn},\cite{kn:mar}.

 Billiards also provide a natural
 arena  for  comparisons  between   classical  and  quantum
 mechanics.  Number   theorical  results   and  conjectures
 (including about  the Riemann  zeta-function) came  out of
 semiclassical studies \cite{kn:gut},\cite{kn:berry}.

 Billiards   whose   boundaries  may   {\em deform}     are   also
 important in  Physics. For  instance, ''Bohr's  liquid drop
 model'' introduced in  the thirties \cite{kn:peis}, assumes  that the
 nucleous is populated by particles,  moving inside it like
 billiard balls. Collisions among these balls are neglected,
 and exchange of energy occurs only between the particles and
 the deforming nucleous  boundary. This  model is  still in
 vogue  and   many   questions   remain   open \cite{kn:abal}.
 The 1-dimensional case already is quite complicated, even if the
 motion of one wall  is prescribed and  the other is fixed;
 this  ''ping-pong'' ball  problem   is  known   as  
''Fermi accelerator'', a basic  model in Plasma  physics 
\cite{kn:fermi},\cite{kn:ll}. By
 registering the velocity change and time at collisions, one
 gets an  area preserving  map  in (energy $\times$ time) space.
 Numerical studies, with different periodic wall functions,
 indicated the existence  of invariant  curves crossing the
 time direction for a whole period, thus limiting the maximum
 energy gain. Rigorous  mathematical  proofs  of the
 existence of such invariant curves were given, in some cases, via KAM
 theory (see, \cite{kn:dou}, 3.IV, and \cite{kn:lae} \S 4).
 Using similar methods, M. Levi has given examples of ''pulsating
 soft circular billiards'' whose energies stay bounded for all
 time (\cite{kn:levi} \S 1).
 We believe it is natural
 to consider ''higher dimensional  Fermi accelerators'', that
 is,  billiards  whose   boundaries  move,   in  particular,
 periodically. 

  The  techniques we  use in  this paper  are elementary.  In
 section~2 we show that a billiard with moving boundary can
 be viewed as a rigid billiard in an augmented phase space.
 The metric,  however,  is {\em singular}.  Section~3 presents
 suitable coordinates,  although  not  canonical,  for  the
 Poincar\'{e} section  of  the  2-dimensional  moving  billiard
 mapping. These coordinates consist on time
 and energy, as in the 1-dimensional Fermi accelerator, and a pair
 ($\varphi,p_\varphi$)
 where  $\varphi$ parametrizes the frozen boundaries.
 In section~4
 we obtain the derivative of the billiard mapping. Section~5
 contains  a  simple  application,  which  hopefully  could
 surprise some  readers.


\section{Time-dependent billiards as rigid billiards in augmented space}

The problem of the time-dependent billiard consists on the study of the billiard
map with a moving boundary.
In this section we show that deforming billiards can be
viewed as the limiting case of {\em rigid} billiards in suitably space, with a 
special {\em singular} metric.
For simplicity we shall restrict ourselves to two dimensional time-dependent
billiard in euclidean space, but everything carries over for
higher dimensions and general riemannian manifolds.

The time dependence of the potential energy $V$ will be indicated by a third
''spatial'' coordinate $z$. The special metric is defined by a generalized kinetic
energy
which depends on a small parameter $\gamma$. (See \cite{kn:arno} \S 19.D)

More precisely, consider a particle in ${\bf R}^3 = (x,y,z)$ moving under the 
potential energy $V(x,y,z)$  and with kinetic energy defined by
\be
T_\gamma (x',y',z') = \ \frac{1}{2} (x'^2 + y'^2 + \frac{1}{\gamma^2} z'^2)
\label{eq:T}
\en
where $\gamma$ is a small parameter,  $'$
stands for $d\, /dt$  and $t$ is a new independent variable.
The equations of motion of this autonomous lagrangian system
are, evidently,
\bea
&(x'',y'') &= \ - {\rm grad}_{(x,y)} V(x,y,z)   \nonumber \\
& z''      &= \ - \gamma^2 {\partial V} / {\partial z}  \ . 
\label{eq:meq}
\ena

Now consider in ${\bf R}^3$ a tube $f(x,y,z)\le 1$, where $f$ is a sufficiently
smooth function, with $\partial f / \partial z$ bounded.
In our setting, we define
\be
V(x,y,z) = \ {\displaystyle\lim_{N\to\infty}} f^N (x,y,z) =
\left\{  
\begin{array} {ll}
0  & \mbox{inside the tube}  \\
\infty & \mbox{outside the tube}
\end{array}
\right.  
\label{eq:V}
\en
( for large but finite $N$ we have a ''soft billiard'').

Therefore we can make computations like in a usual rigid billiard,
using the special metric. (See \cite{kn:corn}, ch.6).

Let $\vec{v}_0$ and $\vec{v}_1$ be the velocity vectors immediately before
and after
the collision of the particle with a point $\vec{q}_1$ of the boundary $f(x,y,z)=1$. Then
\bea
& \vec{v}_1 & = \ \vec{v}_0 - 2 <\vec{v}_0,\vec{n}>_\gamma \vec{n} \\
& \vec{n}   & = \ \frac{\bigtriangledown_\gamma f ( \vec{q}_1)}  
               {\parallel \bigtriangledown_\gamma f   ( \vec{q}_1) 
\parallel_\gamma} \nonumber   
\label{eq:v0n}
\ena
where $<,>_\gamma$ is the riemannian metric given by $T_\gamma$ and 
$\bigtriangledown_\gamma$ its associated gradient operator.

Since
\bean
& \bigtriangledown_\gamma f & = \ \left( \frac{\partial f}{\partial x },
\frac{\partial f}{\partial y },
\gamma^2\frac{\partial f}{\partial z } \right )  \\
& \parallel \bigtriangledown_\gamma f \parallel_\gamma^2 & = \
\left ( \frac{\partial f}{\partial x } \right )^2 +
\left ( \frac{\partial f}{\partial y } \right )^2 +
\gamma^2 \left ( \frac{\partial f}{\partial z } \right )^2 \ 
%\label{eq:delta} 
\enan

It follows that, for $\vec{v}_0 = (a_0, b_0, c_0)$,
\be
\vec{v}_1 = (a_1, b_1, c_1)
 = (a_0, b_0, c_0)
- 2 \frac {
\left ( 
a_0 \frac{\partial f}{\partial x } +
b_0 \frac{\partial f}{\partial y } +
c_0 \frac{\partial f}{\partial z }
 \right )
 }
{
\left ( \frac{\partial f}{\partial x } \right )^2 +
\left ( \frac{\partial f}{\partial y } \right )^2 +
\gamma^2 \left ( \frac{\partial f}{\partial z } \right )^2 
}
\left( \frac{\partial f}{\partial x },
\frac{\partial f}{\partial y },
\gamma^2\frac{\partial f}{\partial z } \right )
\ .
\label{eq:v1}
\en 

\begin{prop}
Take $c_0 = 1$. Then $\displaystyle{\lim_{\gamma \to 0} \vec{v}_1}$ exists 
and is equal to
\be
\vec{v}_1 = 
(a_1, b_1, 1)
 = (a_0, b_0, 1)
- 2 \frac {
\left ( 
a_0 \frac{\partial f}{\partial x } +
b_0 \frac{\partial f}{\partial y } +
\frac{\partial f}{\partial z }
 \right )
 }
{
\left ( \frac{\partial f}{\partial x } \right )^2 +
\left ( \frac{\partial f}{\partial y } \right )^2 
}
\left( \frac{\partial f}{\partial x },
\frac{\partial f}{\partial y },
0 \right )
\ .         
\label{eq:6}
\en
\end{prop} 
\noindent{\bf Proof:} 
Take $c_0=1$ in (\ref{eq:v1}) and then take the limit when $\gamma$ goes to 
$0$.

Observe that proposition~1 tells us that the section $\{z'=1\}$
is invariant by the flow defined by (\ref{eq:meq}), when 
$\gamma \to 0$.


Consider now a time-dependent billiard whose boundary at time $t$ is given by
$C(t)=\{ f(x,y,t)=1 \}$.
Suppose that the particle leaves the boundary 
$C_0= C(t_0)$ at the point $q_0 \in C_0$ with velocity
$\vec{v}_0 = (a_0, b_0)$.
It reaches the boundary again, in some future time $t_1$, at 
a point $q_1 \in C_1 = C(t_1)$, with an exit velocity 
$\vec{v}_1 = (a_1,b_1)$.
If we look the moving boundary as a surface in the 
$(x,y,t)$ space the trajectory of the particle at each instant
will have velocity $\vec{v} = (\dot{x},\dot{y},1)$.
So, the billiard map shall change an initial condition 
$\vec{q}_0 = (q_0,t_0)$ with $\vec{v}_0 = (a_0, b_0, 1)$
into  
$\vec{q}_1 = (q_1,t_1)$ with $\vec{v}_1 = (a_1, b_1, 1)$.
%\begin{figure}
%\vspace{8 truecm}                                                                                      
%\caption{Geometry of the billiard maping.}
%\end{figure}

We will show now that proposition~1  also implies that the time-dependent 
billiard can be thought as the limit case, 
$\gamma \to 0$, of the movement given by (\ref{eq:meq}) and (\ref{eq:V}).


 To  check  that this  is  the  physically correct  formula,
 denote by $\vec{\eta}$  the instantaneous  external normal  vector in
 $xy$-plane, with the usual metric,
 \be
  \vec{\eta}(x,y; t)  =  \frac{(\partial f/\partial x,\partial f/\partial y)}   
           {[(\partial f/\partial x)^2 + (\partial f/\partial y)^2]^{1/2}}                 
 \label{eq:7}
 \en

 Caveat:  do  not  confound $\vec{\eta}$ with $\vec{n}$.  The  latter  is
 3-dimensional, see (\ref{eq:v0n}).

Equation (\ref{eq:6}) is equivalent to 
$$
(a_1,b_1)_{mb} = (a_1,b_1)_{rb}   +   2 \, \frac{-\partial f/\partial t}
{[(\partial f/\partial x)^2 + (\partial f/\partial y)^2]^{1/2}} \,  \vec{\eta}       
%\label{eq:6a}
$$
 where ''mb'' means  ''for our billiard  with moving boundary''
 and ''rb'' means ''for the  instantaneous billiard with rigid
 boundary''. In other words, while $(a_1,b_1)_{rb}$  is given by the
 usual specular reflection,
$$
(a_1,b_1)_{rb} =  (a_0,b_0)   -  2 \,  \frac{(a_0\partial f/\partial x + b_0
\partial f/\partial y)}{(\partial f/\partial x)^2 + (\partial f/\partial y)^2}
 \,   (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})
$$
there is an extra momentum  $2u\vec{\eta}$, with   
$$
  u  =   \frac{-\partial f/ \partial t}
    { [(\partial f/ \partial x)^2 + (\partial f/\partial y)^2]^{1/2}}
$$

 It is easy to see that $u$  is   the
 velocity of the boundary in the direction of the normal $\vec{\eta}$.
 There is a  simple physical explanation for the
 factor 2 in  formulas  (\ref{eq:v1}) and  (\ref{eq:6}).  Usual specular
 reflection law holds if one goes to a reference frame moving
 with constant velocity  $- u\vec{\eta}$ ,  because in  this frame the
 boundary velocity at the point of collision is {\em zero}. The
 normal component of the incoming vector in the new frame is
 equal to the normal  component in the  old frame minus $u\vec{\eta}$.
 Thus, under reflection, the normal component of the velocity
 of the particle gets  an extra $u\vec{\eta}$ term.  Going back to the
 old frame adds one more  $u\vec{\eta}$ .

  We  recall  that  conservation  of  energy  holds  for  any
 rigid billiard; in particular,  using  the  $\gamma$-metric,
$$
  \parallel \vec{v}_1 \parallel ^2_{\gamma} = 
  \parallel \vec{v}_0 \parallel ^2_{\gamma} 
%\label{eq:9}
$$
 or equivalently,
$$
        a_0^2 +b_0^2 +( \frac{1}{\gamma^2})c_0^2 
           = a_1^2 +b_1^2 + (\frac{1}{\gamma^2})c_1^2. 
%\label{eq:9a}
$$

        We are interested in the   $x,y$  part of the energy,
 $$
 2  \Delta E  =  (a_1^2  + b_1^2)  -  (a_0^2  + b_0^2)  
         =  (\frac{1}{\gamma^2})c_0^2 - (\frac{1}{\gamma^2})c_1^2
%\label{eq:10}
$$
where from (\ref{eq:v1})  we have 
$$
c_1 =  c_0 - 2\gamma^2 
\bigl( \frac {\partial f} { \partial z}\bigr)
\bigl(
\frac
{
\partial f/ \partial x \, a_0 + 
 \partial f/ \partial y \, b_0 + 
 \partial f/ \partial z \, c_0 
}{
(\partial f/ \partial x)^2+
(\partial f/ \partial y)^2+
\gamma^2 (\partial f/ \partial z)^2
}
\bigr)
%\label{eq:11}
$$
Therefore,
$$
2 \Delta E_\gamma
=\frac{c_0^2}{\gamma^2}-\frac{1}{\gamma^2}
\left[
c_0 - 
2 \gamma^2
\bigl(
\frac {\partial f} { \partial z}
\bigr)
\bigl(
\frac{
 \partial f / \partial x a_0+ \partial f / \partial y
 b_0 + \partial f / \partial z c_0}
{
(\partial f / \partial x)^2 + (\partial f / \partial y)^2 +
\gamma^2  (\partial f / \partial z)^2
}
\bigr)
\right]^2
%\label{eq:12}
$$
Taking the limit as $\gamma \to 0$ we get

\begin{prop}The  energy change  at  each  collision is
 given by
$$
 \Delta E =
 2 \bigl(
\frac {\partial f} { \partial z}  \bigr)
\bigl(
\frac
{ 
\partial f / \partial x \, a_0 +
\partial f / \partial y\,  b_0 +
\partial f / \partial z
}
{
(\partial f / \partial x)^2 + (\partial f / \partial y)^2 
} 
\bigr)
%\label{eq:12linha}
$$
\end{prop}

{\bf Remark 1:} Although the metric is singular in $\gamma = 0$, if 
$\partial V / \partial z $ (or equivalently $\partial f / \partial z $) 
remains bounded,
then $z''(t) \to 0 $ as $\gamma \to 0$ (see (\ref{eq:meq}) ).
So, if we choose initial conditions such that $z'|_{t=0} = 1 $ and 
$z |_{t=0} = 0$, then
${\lim_{\gamma \to 0} z (t) = t}$, or $z(t)=t+{\cal O}(\gamma ^2)$.     
Indeed, a philosophical principle, well known among physicists, states
that {\em time is just a spatial coordinate whose inertia tends to infinity
(so its motion is uniform)}. We refer to Montgomery \cite{kn:mont} for related
 equations and
bibliography on singular metrics, in the context of Calculus of Variations.

{\bf Remark 2:} Under the change of scale $z=\gamma \zeta$,
 the Lagrangian system (\ref{eq:T}),(\ref{eq:V})
becomes
$T= (1/2)\, (x'^2 + y'^2 + \zeta'^2)$,
$V=V_\gamma= V(x,y,\gamma\zeta)$,
and the singular behavior of the metric as $\gamma \to 0$ is removed.
However, this change is not very useful if the natural initial conditions
indicated above are considered; the velocity component in the $\zeta$ direction
is very large, $d\zeta/dt = 1/\gamma + {\cal O}(\gamma)$ and it will remain 
as such, at least
for $t = {\cal O}(1)$. 

{\bf Remark 3:} At first  sight,  the small  parameter
$\gamma$
 seems to be artificial. However, taking $\gamma = i/c $,
(\ref{eq:V}) becomes the
 Lorentz's metric. Relativistic billiards may turn out to be
 good models for  highly accelerated  particles confined on
 magnetic bottles (this, incidentally, was Fermi's initial
 motivation).


\section{Poincar\'e's  integral invariant: coordinates for
 a section}

Under  Legendre's
transformation $p_x  =  x'$, $ p_y =  y'$, $ p_z  =  (1/\gamma^2)z'$, our
autonomous Lagrangian of mechanical type (\ref{eq:T}),(\ref{eq:meq}) 
turns into the Hamiltonian
\be
         H = \frac{1}{2} [ p_x^2 + p_y^2 + \gamma^2p_z^2] + V(x,y,z) .
\label{eq:13}
\en

 Observe  that  since $z'  =  1  + {\cal O}(\gamma^2)$,
 $ p_z  = {\cal O}(1/\gamma^2)$  and
 $\gamma^2p_z^2 = {\cal O}(1/\gamma^2)$. Hence  the term $\gamma^2p_z^2$
 cannot be dropped
 when $\gamma$ goes to zero. One way to circumnvent this difficulty is to
 consider the Hamiltonian
 $H =  (1/2)  [ p_x^2  +  p_y^2]  + V(x,y,z)$  together  with the
 {\em constraint} $z'  = 1$ in the context of Dirac's theory.
 For  our purposes  it is  simpler to
 rewrite (\ref{eq:13}) as
\bean
   & H & = E  + \frac{1}{2\gamma^2}  z'^2 ,   \\
   & E &=  \frac{1}{2} \, [  p_x^2 +  p_y^2 ]  + V(x,y,z), 
%\label{eq:13'}
\enan   
with E being the (relevant) finite part of the energy when
$\gamma$
is infinitesimal.  Differentiating, we  get
$$
    dH  = dE  +  {z'}  (1/ \gamma^2) dz'  = dE  + z'dp_z \  , 
    \qquad z'  = 1 + O(\gamma^2),
%\label{eq:14}   
$$

  If we confine  ourselves to a  fixed total energy manifold,  then
 $dH=0 $  and
\be
        dp_z = -dE/z' = -dE + O(\gamma^2).
\label{eq:15}   
\en

   We  want   to  construct   suitable  coordinates  for   the
 moving billiard mapping, describing consecutive phase space
 points immediately  after  collision,  and  calculate  the
 associated Poincar\'{e}'s integral  invariant (see \cite{kn:arno} \S 38).
  This   mapping  is 
 {\em four}
 dimensional: fixing the total energy $H$ in (\ref{eq:13}), one variable
 is cut down, so the flow  is restricted to a 5-dimensional
 manifold;   the   successive    collisions   represent   a
 4-dimensional submanifold transversal to the flow.

  We change  the cartesian  coordinates $x,y,z$ to  generalized
 coordinates $\zeta,\varphi,\xi$  as  follows:  
 let  $x_0  = \alpha (z,\varphi)$,
 $ y_0 = \beta (z,\varphi)$  represent parametrically  the moving  boundary
 $f(x,y,z)=1$. We dont need to assume that the $z$-curves are orthogonal to
 the $\varphi$-curves. Let $\xi$ describe  outward motion along normals
 (with frozen $z$).  The desired  coordinate transformation 
 $X  :  (\zeta ,\varphi,\xi )\longrightarrow     (x,y,z)$ is
\bean 
x  = &x(z,\varphi,\xi)  & =\alpha(z,\varphi)
       +\xi\,\frac{\partial\beta/\partial\varphi}
 {[(\partial\alpha/\partial\varphi)^2 
+ (\partial\beta/\partial\varphi)^2]^{1/2}} 
  \\
y  = &y(z,\varphi,\xi)  & =\beta(z,\varphi)
       -\xi\,\frac{\partial\alpha/\partial\varphi}
       {[(\partial\alpha/\partial\varphi)^2 
+ (\partial\beta/\partial\varphi)^2]^{1/2}}  \\
z  = &\gamma\zeta &
%\label{eq:16}
\enan
Note that $\xi =0$ means that we are on the moving boundary.   
The kinetic energy T will be given by    
$$
 T  = (1/2) (\zeta', \varphi', \xi') G (\zeta', \varphi', \xi')^{tr}
$$
with $G = A^{tr}  A$, where $ A^{tr}$ is  
the matrix whose  rows are $X_\zeta,X_\varphi,X_\xi$.
Moreover, Legendre's transformation yields
$$
 (p_\zeta,p_\varphi,p_\xi) = G(\zeta',\varphi',\xi')^{tr}
   =  A^{tr} A A^{-1}(x',y',z')^{tr}
   =  A^{tr} (x',y',z')^{tr}
$$
Thus for instance, $p_\varphi$ is equal to the scalar product  of  $X_\varphi$  
 with
$(x',y',z')$.  In our case, since  $z_\varphi \equiv 0$,  we have
\be
             p_\varphi  = x' x_\varphi + y' y_\varphi
\label{eq:17}
\en
   Poincar\'e's  1-form  writes,   via  pullback  to  the   new
 coordinates, as
$$
\lambda  =  x' (  x_\varphi d\varphi  + x _\xi d\xi +  x_z dz  ) +
            y' (  y_\varphi d\varphi  + y _\xi d\xi +  y_z dz ) + p_z dz
$$
Our  4-dimensional  section  is  obtained  by  fixing   the
total energy in (\ref{eq:13}) and taking  $\xi= 0$ . So
$\lambda$   simplifies into
$$
\lambda =  (x' x_\varphi + y' y_\varphi) d\varphi +  (x' x_z + y'y_z+ p_z) dz
$$
or, in view of (\ref{eq:17}),
$$
\lambda    =   p_\varphi  d\varphi  +   (F+p_z)dz
%\label{eq:18}
$$
where we have defined  $F   =  x'x_z  + y'y_z$.

Finally, from (\ref{eq:15}) and letting  $\gamma \to 0$, we get:
\begin{prop}
The  conserved  2-form  for  the  Poincar\'e section is
\be
\omega     =  dp_\varphi d\varphi  -  dt d(E-F) \ .
\label{eq:19}
\en
\end{prop}

  Although  the pairs  $(\varphi,  p_\varphi)$,  $(E-F,  t)$ are  canonically
 conjugate, physically  it  would be  more  natural  to map
 over $((\varphi, p_\varphi)  , (E,  t))$. {\em Helas}, the  quantity $F$  is an
 unavoidable nuisance, inherited from the  time-dependence.

    Perhaps   it    is   instructive    to   look   into    the
 1-dimensional case, where  the motion  is confined  to the
 $x$--axis and there are elastic collisions with  moving
 walls $x=a(t)$ and $ x =b(t) $. Let's  look at  a collision with
 the left wall. The first term in (\ref{eq:19}) does not appear, and
 the second gives
\be
\omega 
=- dt \, d \left[ \frac{1}{2} (x')^2 - x' a_t \right] 
=  dt ( -   x'   dx'    +    a_t   dx'    +    x'   a_{tt}    dt ) 
= ( a_t - x') dt dx' .
\label{eq:20}       \ 
\en

  As a motivation  for the next  section, we present now  the
 derivative of the mapping which gives successive collisions
 with the two moving walls. This is obtained by an elementary
 calculation. Suppose the particle leaves  the left wall at
 time $t_0$  with  velocity  $x'_0$ ,  and  that  it  reaches the
 right wall at  a later time  $t_{1/2}$ , given  by the implicit
 equation
\be
    b(t_{1/2})= a(t_0)+x'_0 (t_{1/2}-t_0)  \ .
\label{eq:21}
\en
After the collision we have
\be
 x'_{1/2} = -x'_0 + 2 \frac{db}{dt}(t_{1/2}) \ .
\label{eq:22}
\en
Analogously, we get
\bean
&a(t_1)  &=  b(t_{1/2})  +  x'_{1/2}  (t_1  -t_{1/2}) \ ,
%\label{eq:23} 
 \\
&x'_1    &= -  x'_{1/2} + 2\frac{d a}{dt}(t_1) \ . 
\enan
Differentiating (\ref{eq:21}),(\ref{eq:22}) we obtain
\begin{eqnarray*}
& b'(t_{1/2})   dt_{1/2} & =   a'(t_0)dt_0   +  x'_0   (dt_{1/2}  -   dt_0)
 + dx'_0 (t_{1/2} - t_0) \\
& dx'_{1/2} &= -dx'_0 + 2b''(t_{1/2}) dt_{1/2}
\end{eqnarray*}
so that,  after a simple algebra,
$$
                  (dt_{1/2} , dx'_{1/2})^{tr} =  B (dt_0 , dx'_0)^{tr}
$$
where
\be
B= \left(
\begin{array}{cc}
(a'_0   -    x'_0)/(b'_{1/2}     -    x'_0)
& (t_{1/2} - t_0)/(b'_{1/2} - x'_0)
 \\
  2b''_{1/2}  (a'_0-x'_0)/(b'_{1/2}-x'_0)
&     -1          +   2b''_{1/2} (t_{1/2} - t_0)/(b'_{1/2} - x'_0)
\end{array}
\right )
\label{eq:24}
\en
 Similarly  we  can  write
 $(dt_1  ,  dx'_1)^{tr}  = A  (dt_{1/2}  ,  dx'_{1/2})^{tr}$
with
$$
A= 
\left(
\begin{array}{cc}
 ( b'_{1/2}   -   x'_{1/2} ) / ( a'_1  -    x'_{1/2} )
& (t_1 - t_{1/2}) / (a'_1  - x'_{1/2})
\\
2a''_1 ( b'_{1/2} -    x'_{1/2} ) / (a'_1-x'_{1/2} ) 
&    -1    +   2a''_1 ( t_1-t_{1/2} ) /( a'_1-x_{'1/2})
\end{array}
\right)
$$

 Now we can present  a  direct verification  of the  invariance of  the
 measure    on the plane  $(x,t)$  given by (\ref{eq:20}).  Since
$$
   (b'_{1/2} - x'_0) =  - (x'_{1/2} - b'_{1/2})
$$
then
\bean
&&det  B   =  -  (a'_0   -  x'_0)/(b'_{1/2}  -   x'_0)
          =     (a'_0   -  x'_0)/(x'_{1/2}  -  b'_{1/2})  ,
%\label{eq:25} 
\\
&& (a'_0    -   x'_0)    dt_0    dx'_0   =    (x'_{1/2}   -    b'_{1/2})
 dt_{1/2} dx'_{1/2} 
\enan
 as desired.

\section{ Derivative of the moving billiard mapping. Case
 n=2}

  Inspired  by  the  last  calculation,  we  now compute  the
 derivative  of  the  time-dependent  billiard  mapping  in  two
 dimensions. The relevant variables are depicted in Fig.1.
 We  assume  for  convenience  that  the  curves $ C(t)$   are
 convex;  following  Berry  \cite{kn:berry2}  we  use 
 the angle  $\varphi$ with a fixed direction as parameter,
 $\varphi \in [0,2\pi)$. If $C(t)$  ceases to be  convex,
 one  may pass to  arc length $ s \in [0,l)$.
\begin{figure}
\vspace{8 truecm}                                                                                      
\caption{}
\end{figure}

 Suppose the particle leaves  the boundary $C_0  = C(t_0)$ from
 point ${q}_0$ with  velocity $\vec{v}_0$ ,  making an angle
 $\alpha_0$ with the
 instantaneous tangent vector 
$\vec{\tau}_0=(\cos \varphi_0,\sin \varphi_0)$.
 It reaches the  boundary again  in  some future  time  $t_1$, at
 point ${q}_1 \in  C_1 = C(t_1)$,  where the exit  velocity $\vec{v}_1$
 makes angle $\alpha_1$ with the vector $\vec{\tau}_1 = 
 (\cos \varphi_1, \sin \varphi_1)$. We denote
 by $\alpha_1^{\ast}$  the {\em entry}  angle  at ${q}_1$  .
 In general $\alpha_1^{\ast} \neq \alpha_1$.
 Elementary geometry yields  (observe that no  harm is done
 from  $C_0$ and $C_1$ being different curves):
$$
 \alpha_0  +  \alpha_1^{\ast} = \varphi_1 - \varphi_0 
\qquad
\hbox{(hence $d \alpha_1^{\ast}  = -d\alpha_0 - d\varphi_0 + d\varphi_1$)}
%\label{eq:26}
$$
Given  ${q}_0 \in C(t_0)$, ${q}_1 \in C(t_1)$,  clearly
$$
 \parallel \vec{v}_0 \parallel =  (2E_0)^{1/2} =
  \frac{ \parallel {q}_1 - {q}_0 \|}{t_{01}} \quad   ,
\quad t_{01} =  t_1 - t_0  \ .
%\label{eq:27}
$$
   If  the  boundary  is $ C(t):  f(x,y,t)=1$,  then  given 
 ${q}_0 \in  C(t_0)$ and $\vec{v}_0$ , in order to obtain $t_{01}$
 (and hence ${q}_1 = {q}_0 + t_{01} \vec{v}_0$ )
   we must solve the implicit equation (for   $t_{01}$)
$$
       f( {q}_0 + t_{01} \vec{v}_0 ,  t_0 + t_{01}) = 1 ,
%\label{eq:28}
$$

\begin{lem}
Let $u_1$ be the velocity of the boundary at $({q}_1,t_1)$
 in the direction of  the exterior normal  and $R$ the radius
 of curvature of  the instantaneus  boundary at  the impact
 point. Then
\bea
&& p_\varphi = (2E)^{1/2} R \cos \alpha \nonumber \\
&& \cos \alpha_1   =    (E_0/E_1)^{1/2} \cos \alpha_1^\ast   \label{eq:29}\\
&& E_1   =  E_0   +  2u_1 [u_1 - (2E_0)^{1/2} \sin \alpha_1^\ast] \ .\nonumber
\ena
\end{lem}
\noindent{\bf Proof:} 
 From (\ref{eq:17}) we get 
$p_\varphi  = (x_\varphi,y_\varphi) \cdot (x',y') =
(\partial s/\partial \varphi)  \vec{\tau} \cdot \vec{v}$.
 The  boundary is  parametrized  by  
$(x,y) =  (x(\varphi,t), y(\varphi,t))$,
$R  =  \partial s / \partial \varphi$ is the radius of curvature
and $\vec{\tau}$  is  the  instantaneous tangent
vector.
 There is no  ambiguity by taking  the velocity $\vec{v} =
 (x',y')$ either immediately before or after impact, because
 the tangential  component does  not change.  In particular
$ (E_0)^{1/2} \cos \alpha_1^\ast = (E_1)^{1/2} \cos \alpha_1$.
 Proposition~2  gives 
 $E_1 =
 E_0 + 2  (f_t/ \| \nabla f \|)
( f_t/ \| \nabla f \|  +  \vec{v}_0 \cdot \nabla f / \| \nabla f \|)$.
 Recall
 also (see  (\ref{eq:7}))  that
$  \vec{\eta}  =  \nabla f / \| \nabla f \|$
 is  the  exterior   normal and 
$u  = -f_t/ \| \nabla f \| $. Finally,  observe from Fig.1 that 
$ \sin \alpha_1^\ast  = \vec{\eta}_1 \cdot \vec{v}_0 / \| \vec{v}_0 \|$.
Formulas (\ref{eq:29}) follow.

 Now   we   describe   the   derivative   of  the   billiard
 mapping
$$
(  \varphi_0   , \alpha_0   ,   E_0  ,   t_0   ) \rightarrow
(  \varphi_1   , \alpha_1   , E_1 , t_1 )
$$
 where, in order to compare results with rigid billiards, we
 have  replaced   the   variable  $p_\varphi$   by $\alpha$.
  A  tedious
 calculation is outlined on the Appendix:

%\pagebreak
\begin{prop} (''bank  of  differentials''  for  the  $4\times 4$
jacobian).

\bigskip

i) $\qquad \qquad dt_1= C_{41} d\varphi_0 + C_{42} d\alpha_0 
+ C_{43}dE_0 + C_{44}dt_0$
\begin{eqnarray*}
&&C_{41} [-\sin\alpha_1^\ast(2E_0)^{1/2}+u_1]
=-t_{01}\cos\alpha_1^\ast(2E_0)^{1/2}-R_0 \sin (\varphi_0-\varphi_1)
\\
&&
C_{42}[-\sin\alpha_1^\ast(2E_0)^{1/2}+u_1]
=-t_{01}(2E_0)^{1/2}\cos\alpha_1^\ast
\\
&&
C_{43}[-\sin\alpha_1^\ast(2E_0)^{1/2}+u_1]
=t_{01}\sin\alpha_1^\ast/(2E_0)^{1/2}
\\
&&
C_{44}[-\sin\alpha_1^\ast(2E_0)^{1/2}+u_1]
=-\sin\alpha_1^\ast(2E_0)^{1/2}+u_0
 \cos (\varphi_0 - \varphi_1 )
\end{eqnarray*}

\bigskip

ii) 
$ \qquad \qquad
d\varphi_1  =  C_{11}   d\varphi_0  
+  C_{12}   d\alpha_0  +  C_{13}   dE_0  +  C_{14}  dt_0
$
\begin{eqnarray*}
&&R_1  C_{11}  = 
R_0  \cos (\varphi_0  -  \varphi_1)  +  t_{01}  \sin \alpha_1^\ast(2E_0)^{1/2}  +
 (2E_0)^{1/2} \cos \alpha_1^\ast C_{41}
\\
&&
R_1 C_{12} =
t_{01} \sin \alpha_1^\ast(2E_0)^{1/2} + (2E_0)^{1/2} \cos \alpha_1^\ast C_{42}
\\
&&
 R_1 C_{13} =
 t_{01} \cos \alpha_1^\ast/(2E_0)^{1/2} 
+ (2E_0)^{1/2} \cos \alpha_1^\ast C_{43}
\\
&&
  R_1  C_{14}  =
  u_0  \sin (\varphi_0 - \varphi_1)  
+  (2E_0)^{1/2}  \cos \alpha_ 1^\ast (-1  + C_{44})
\end{eqnarray*}

\bigskip

iii)
$ \qquad \qquad
  dE_1  =  C_{31}  d\varphi_0   +  C_{32}  d\alpha_0  
+   C_{33}  dE_0  +  C_{34}
 dt_0 
$
\begin{eqnarray*}
C_{31} =&&2C_{11}\partial  u_1/\partial \varphi_1 
[ 2u_1  -  \sin \alpha_1^\ast (2E_0)^{1/2} ]  
+\\
&+&  2C_{41} \partial  u_1/ \partial  t_1 [ 2u_1 
- \sin \alpha_1^\ast (2E_0)^{1/2}]
+\\
&+ &2u_1 \cos \alpha_1^\ast (1 - C_{11})(2E_0)^{1/2}
\\
C_{32} = &&2C_{12} \partial  u_1/ \partial \varphi_1 
  [ 2u_1  -  \sin \alpha_1^\ast (2E_0)^{1/2} ]  
  +
\\ &+& 2C_{42} \partial u_1/\partial t_1 [ 2u_1 
- \sin \alpha_1^\ast(2E_0)^{1/2} ]
+\\
& + &2u_1 \cos \alpha_1^\ast (1 - C_{12})(2E_0)^{1/2} 
\\
  C_{33} = &&1 + 2C_{13} \partial  u_1/\partial \varphi_1 
  [ 2u_1 - \sin \alpha_1^\ast  (2E_0)^{1/2}  ]
  +\\
&+& 2C_{43} \partial u_1/\partial t_1 [ 2u_1 
- \sin \alpha_1^\ast (2E_0)^{1/2}]
-\\
&  - &2u_1 [ \cos\alpha_1^\ast C_{13}(2E_0)^{1/2} 
+ \sin \alpha_1^\ast/(2E_0)^{1/2}]
\\
  C_{34} = &&2C_{14} \partial  u_1/\partial \varphi_1 
  [ 2u_1 - \sin \alpha_1^\ast (2E_0)^{1/2} ]  
  + 
\\
&+&2C_{44} \partial  u_1/\partial  t_1 [ 2u_1 -
 \sin \alpha_1^\ast(2E_0)^{1/2}]
-\\
& - &2u_1\cos \alpha_1^\ast C_{14} (2E_0)^{1/2} 
\end{eqnarray*}

\bigskip

iv) 
$ \qquad \qquad                                                                            
 d\alpha_1 = C_{21} d\varphi_0 + C_{22} d\alpha_0 
+ C_{23} dE_0 + C_{24} dt_0 
$
\begin{eqnarray*}
&&-\sin \alpha_1 C_{21} =(E_0/E_1)^{1/2} [ \sin \alpha_1^\ast (1- C_{11})
  - C_{31} \cos \alpha_1^\ast /(2 E_1)] 
\\
&&
  -\sin \alpha_1 C_{22} =  (E_0/E_1)^{1/2} [ \sin \alpha_1^\ast(1- C_{12})
  - C_{32} \cos \alpha_1^\ast /(2 E_1)] 
\\
&&
  -\sin \alpha_1 C_{23} = -(E_0/E_1)^{1/2}\sin \alpha_1^\ast C_{13} 
  + \cos \alpha_1^\ast /(2(E_0 E_1)^{1/2}) - \\
&& \qquad \qquad  \ \ \ \ \ \ \
- C_{33}(E_0/E_1)^{1/2}
  \cos \alpha_1^\ast /(2E_1)  
\\ 
&&
  -\sin \alpha_1 C_{24} = (E_0/E_1)^{1/2} [ - \sin \alpha_1^\ast C_{14} 
  - C_{34} \cos \alpha_1^\ast /(2 E_1)]
\end{eqnarray*}
 (Observe    the    "telescopic"    structure    of    these
 formulas)
\end{prop}

\section{ Example: loss of stability of the energy component}

   In  this   section  we   consider  the  simplest   possible
 situation, where at the points of contact
$$
  \alpha_0 = \alpha_1 = \pi/2 \quad ,\quad
  \varphi_1 - \varphi_0 = \pi \quad , \quad
   u_1 = u_0 = 0 \ .
%\label{eq:30}
$$             
 i.e, we  have a  2-bounce periodic  orbit whose  energy is
 unchanged  at  the  collisions.  First   we  look  at  the
 ''geometrical block''  $A$  associated  to  the  corresponding
 rigid  billiard   mapping $(\varphi_0, \alpha_0) \to 
 (\varphi_1, \alpha_1)$.  Proposition~5 yields 
 (compare with Berry \cite{kn:berry2})
$$
 A=  \left(
\begin{array}{cc}
 C_{11} & C_{12}
\\
 C_{21} & C_{22}                                  
\end{array}
\right )
$$ 
\bean
&& C_{11} = -R_0/R_1 + t_{01}(2E_0)^{1/2}/R_1  \\
&& C_{12} =  t_{01}(2E_0)^{1/2}/R_1         \\
&& C_{21} = -1 + C_{11}                      \\
&& C_{22} = -1 + C_{12}                       
%\label{eq:31}
\enan

 We  take for  simplicity $R_0=R_1$ and  choose  units so  that
 $2E_0 =1$  and  $t_{01} =  1$.  The distance  between  the contact
 points is  $d = t_{01}(2E_0)^{1/2} = 1$ so
$$
 A=  \left(
\begin{array}{cc}
 2a + 1 & 2a + 2
\\
 2a & 2a +1                                  
\end{array}
\right )
\quad \hbox{,where} \quad
 a = (1/2R) - 1 > -1
%\label{eq:32}
$$
  The  stability  index  is $\chi_A = tr(A)/2 = 2a + 1 = (1/R)-1$, 
 so the orbit is unstable for  $R < 1/2$, that is, if the
 osculating circles do not touch (these facts are well known:
 a two periodic orbit is linearly stable if and only if $ t_{01} < R_0 + R_1$
 and $(t_{01} - R_0)(t_{01} - R_1)>0$; see \cite{kn:berry2} \S 3
 and \cite{kn:woj}).
  Next  we   look  at  the   ''Fermi  block''  $B$,
 associated, by  restriction,  to  the  1-dimensional Fermi
 accelerator $(E_0,t_0) \to (E_1,t_1)$. From  proposition~5 we have
 (equivalently, it follows from (\ref{eq:24}) )
$$
 B=  \left(
\begin{array}{cc}
 C_{33} & C_{34}
\\
 C_{43} & C_{44}                                  
\end{array}
\right )
$$
\bean
&& C_{33} = 1 - 2(\partial u_1/\partial t_1)t_{01}/(2E_0)^{1/2}%\label{eq:33}
\\ 
&& C_{34} = -2(\partial u_1/\partial t_1)(2E_0)^{1/2}  \\
&& C_{43} =  t_{01}/2E_0        \\
&& C_{44} = 1    
\enan
so that, since  $t_{01} = 1$, $2E_0 =1$,
$$
 B=  \left(
\begin{array}{cc}
 -2b + 1 & -2b
\\
 1 & 1                                  
\end{array}
\right )
\quad \hbox{,where} \quad
 b = \partial u_1/ \partial t_1
%\label{eq:34}
$$
  The stability index is  $\chi_B = tr(B)/2 = -b +1$ : the orbit  is
 stable for  $0< b < 2$.

 To illustrate the loss  of stability due to the  underlying
 2-dimensional geometry, we take the following values for the
 parameters, yielding  the saddle-center case:
\begin{enumerate}
\item
$b = 1$,  so   the  eigenvalues   of $ B$   are $ i,-i$
 (most stable case for $B$) .
\item
$a = 1/8$, so for $A$ we have $\lambda_1=2 , v_1 =  (3,1) ;
 \lambda_2 = 1/2 , v_2 = (-3,1)$.
\end{enumerate}

 Due  to  the  symmetry  $R_0=R_1$  ,  the  derivative  of   the
 second bounce  mapping  corresponds  to a matrix  $C^2$, where $C$
is
$$
C=
\left (
\begin{array}
{cc}
A & \begin{array}{cc} \ \ 0\ & \ \ 0\ \\ 
                      \ \ 0\ & \ \ 0\ 
    \end{array} 
\\
\begin{array}{cc}    C_{31} &  C_{32} \\                                      
                      0     &  0
\end{array}
&
B
\end{array}
\right )
$$
\bean
&& C_{31}   =    -(5/2) \partial u_1/\partial \varphi_1   ,%\label{eq:35}
\\
&& C_{32}   =   -(9/2)  \partial u_1/\partial \varphi_1 . 
\enan
 
The full eigenvector corresponding to $\lambda_1 =  2$  is
$$
   v = ( 3 , 1 , -(12/5)\partial u_1/\partial \varphi_1 ,
         (4/5)\partial u_1/\partial \varphi_1 ).
%\label{eq:36}
$$
  
{\em The  presence  of  the coupling  terms  $C_{31},  C_{32}$ implies
 linear loss of stability for the energy. This effect would not be
 seen if  we  looked  only  to  the  1-dimensional  Fermi accelerator
 obtained by restricting the dynamics  to  the  line 
$ {q}_0 {q}_1$}.


\appendix
\section*{Appendix: The jacobian matrix:  proof of proposition 5}

 i) Calculation of $dt_1$.

   Linearizing  $f({q}_0 + d{q}_0, t_0 + dt_0) = 1  ,
  f({q}_1 + d{q}_1, t_1 + dt_1) = 1$  we get
\be
 \bigtriangledown f_{({q}_0,t_0)}\cdot d{q}_0  +
   (\partial f/\partial t)_{({q}_0,t_0)}dt_0  =
 \bigtriangledown f_{({q}_1,t_1)}\cdot d{q}_1  +
   (\partial f/\partial t)_{({q}_1,t_1)}dt_1  
\label{eq:a1}
\en
   Clearly, ${q}_0 + t_{01} \vec{v}_0= {q}_1$
 implies to first order
\be
 d{q}_0 + t_{01} d \vec{v}_0 + (dt_1 - dt_0) \vec{v}_0
   = d {q}_1.
\label{eq:a2}
\en
   Multiplying \ref{eq:a2} by $\vec{\eta}_1 = 
 \bigtriangledown f({q}_1,t_1)/
  \parallel \bigtriangledown f({q}_1,t_1)\parallel$
 we obtain
\be
  dt_1( u_1 - \vec{v}_0 \cdot \vec{\eta}_1 ) = 
  -dt_0 \vec{v}_0 \cdot \vec{\eta}_1 
  + t_{01} d\vec{v}_0 \cdot \vec{\eta}_1
  + d{q}_0 \cdot \vec{\eta}_1
\label{eq:a3}
\en
  We  must express  the  scalar products  in  (\ref{eq:a3}) in 
 terms  of the geometrical variables  of Fig.1. Clearly 
 $d{q}_0 = d\varphi_0 R_0 \tau_0 + d\xi_0 \vec{\eta}_0$
 where, because  of (\ref{eq:a1}), \break
 $d\xi_0 =u_0dt_0$
 so that
\be
  d{q}_0 = d\varphi_0 R_0 \vec{\tau}_0 + u_0 dt_0 \vec{\eta}_0  
 \qquad \qquad (d{q}_1  
= d\varphi_1 R_1 \vec{\tau}_1 + u_1 dt_1 \vec{\eta}_1)
\label{eq:a4}                                                                      
\en
>From  $\vec{v}_0 = (2E_0)^{1/2}[\cos(\alpha_0)\vec{\tau}_0
  - \sin(\alpha_0)\vec{\eta}_0]$   we get
\bean
d\vec{v}_0 
= && [dE_0/(2E_0)^{1/2}][\cos (\alpha_0) \vec{\tau}_0 
- \sin(\alpha_0) \vec{\eta}_0 ]
+ \\
 &+& (2E_0)^{1/2} [ -\sin(\alpha_0)
  \vec{\tau}_0 + \cos(\alpha_0)\vec{\eta}_0] d\alpha_0 +
%\label{eq:a5}
\\
&+&    (2E_0)^{1/2} [ \cos(\alpha_0)d \vec{\tau}_0 
+ \sin(\alpha_0)d\vec{\eta}_0] 
\enan
where $\vec{\tau}_0 = (\cos\varphi_0 , \sin\varphi_0)$,
$\vec{\eta}_0 = (\sin\varphi_0 , -\cos\varphi_0 )$.
Finally,  after  a   few  manipulations  we  get   (observe
 Fig.1)
\bea
&\vec{v}_0 \cdot \vec{\eta}_1 &= 
 \sin\alpha_1^{\ast} (2E_0)^{1/2}\nonumber\\
& d{q}_0 \cdot \vec{\eta}_1 &= 
-d\varphi_0 R_0 \sin(\varphi_0-\varphi_1)
  +  dt_0u_0\cos(\varphi_0-\varphi_1) \label{eq:a6} \\
&d\vec{v}_0 \cdot \vec{\eta}_1 &= 
[dE_0/(2E_0)^{1/2}] \sin\alpha_1^{\ast} -
 (2E_0)^{1/2} \cos\alpha_1^{\ast} [ d\alpha_0 + d\varphi_0]\ . \nonumber
\ena
 Substituting  (\ref{eq:a6}) into (\ref{eq:a3}) we  obtain  the   desired
 formula for $dt_1$.

 ii)  Calculation of $d\varphi_1$  .

    From (\ref{eq:a2}) and (\ref{eq:a4}) we have
$$
  R_1d\varphi_1 = d{q}_1 \cdot \vec{\tau}_1 =
  [d{q}_0 + t_{01}d\vec{v}_0 + (dt_1 - dt_0)\vec{v}_0] \cdot \vec{\tau}_1
%\label{eq:a7}
$$
 and analogously to (\ref{eq:a6})
\bean
&\vec{v}_0 \cdot \vec{\tau}_1 &= \cos\alpha_1^{\ast} (2E_0)^{1/2}\\
& d{q}_0 \cdot \vec{\tau}_1 &= d\varphi_0 R_0 \cos(\varphi_0-\varphi_1)
  +  dt_0u_0 \sin(\varphi_0-\varphi_1) %\label{eq:a8}
 \\
&d\vec{v}_0 \cdot \vec{\tau}_1 &= [dE_0/(2E_0)^{1/2}] \cos\alpha_1^{\ast} +
 (2E_0)^{1/2} \sin\alpha_1^{\ast} [ d\alpha_0 + d\varphi_0] 
\enan
 from which the formula for $d\varphi_1$   follows.

 iii)   Calculation of  $dE_1$  and $d\alpha_1$.

Recall that (lemma~4) 
\begin{eqnarray*}
&& E_1 = E_0 + 2u_1^2 - 2u_1 \sin\alpha_1^{\ast}(2E_0)^{1/2}, \\ 
&& \cos \alpha_1 = \cos \alpha_1^{\ast}(E_0/E_1)^{1/2} .
\end{eqnarray*}
The    rest    of   proposition~5    is   obtained    by
 differentiating these expressions, using  that $d\alpha_1^{\ast} =
 -d\alpha_0 - d\varphi_0  +  d\varphi_1$,  and  inserting  the 
 previously computed  differentials for $dt_1$ and $d\varphi_1$ .

\noindent
{\bf Acknowledgements.}  We  thank  Prof.  Raul  Donangelo (IF, UFRJ) for
 informing us about the liquid drop model in nuclear physics and
 Prof. Nicolau  Cor\c c\~ao Saldanha (Departamento de Matem\'atica, PUC-RJ)
 for showing us possible connections
 with  relativistic billiards.
 We thank SBM (Sociedade  Brasileira de Matem\' atica), FAPEMIG
(Funda\c c\~ao de Amparo \`a Pesquisa do Estado de Minas Gerais)
and PrPq-UFMG (Pr\'o-Reitoria de Pesquisa) for sponsoring  visits of two of us
(JK and RM) to the Universidade Federal de Minas Gerais.



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 Physical Review C {\bf 46:1}, 380-381 (1992) \\
 see also
 J.Warnbach: Nuclear sound, Contemp.Phys. {\bf 32}, 291-304 (1991).
\bibitem{kn:arno}
 V.I.Arnold: {\em Mathematical Methods of Classical Mechanics},
 Springer-Verlag,
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\end{thebibliography}

\vskip 2.5 truecm
{
\small
{\sc Jair Koiller}\\
Laborat\'orio Nacional de Computa\c c\~ao Cient\'\i fica\\
R. Lauro Muller 455\\
22290 Rio de Janeiro, Brasil.\\
email:userjako@lncc.bitnet

{\sc Roberto Markarian}\\
IMERL, Facultad de Ingenier\'\i a \\
C.C. $N^{\underline {\rm o}}$ 30 \\
Montevideo, Uruguay.\\
email:roma@iie.edu.uy

{\sc Sylvie Oliffson Kamphorst}\\
Departamento de Matem\'atica, ICEx, UFMG\\
C.P. 702\\
30161 Belo Horizonte, Brasil.\\
email:syok@brufmg.bitnet     

{\sc S\^onia Pinto de Carvalho}\\
Departamento de Matem\'atica, ICEx, UFMG\\
C.P. 702\\
30161 Belo Horizonte, Brasil.\\
email:icedo@brufmg.bitnet 
}

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