\documentclass[12pt]{article}
\usepackage{a4}
\usepackage[short,nodayofweek]{datetime}
\usepackage{graphicx}
\usepackage{bm}
\usepackage[colorlinks=true,
            linkcolor=blue,
            urlcolor=blue,
            citecolor=gray]{hyperref}
\usepackage{multirow}
\usepackage{booktabs}

\usepackage{fancyhdr}
\setlength{\headheight}{15.2pt}
\renewcommand{\headrulewidth}{0pt}
\pagestyle{fancyplain}
\fancyhf{}
\rhead{ \fancyplain{}{Version: \today\,\,\currenttime; svn-65} }
\cfoot{ \fancyplain{}{\thepage} }


\renewcommand\textfraction{0.1}
\setlength{\parindent}{1em}

\renewcommand{\vec}[1]{{\boldsymbol{\mathbf{#1}}}}
\newcommand{\tensor}[1]{\underline{\underline{\boldsymbol{\mathbf{#1}}}}}

\begin{document}
\noindent{\Huge\sf Lecture notes for PC2132\\[5mm](Classical Mechanics)}\vspace*{1cm}

\noindent{\large\sf For AY2017/18, Semester 1\\[5mm]Lecturer: Christian Kurtsiefer,
  Tutors: Adrian Nugraha Utama, Shi Yicheng, Jiaan Qi, Tan Zong Sheng}\vspace*{1cm}

\section*{Disclaimer}
These notes are by no means a suitable replacement for a proper textbook on
classical  mechanics, nor should it replace your own notes.
% There is no promise to have them ready or complete at a well-defined time.
It is just a best-effort affair with the intention to be useful.

This is a document in process --  it hopefully
does not contain too many mistakes, but please contact us if you
feel that you spotted one.\vspace*{1cm}

%\noindent State of this document: \today\,\,\currenttime; svn-55
\vspace*{2cm}


\section*{Notation}
There is an attempt to do use consistent notations through this lecture. Below
is a list what symbols typically refer to, unless they are referenced to
otherwise.\vspace*{1em}

\begin{tabular}{ll}
$\vec{a}$& a vector; often the acceleration \\
$\vec{e}$& a unit vector, i.e., a vector of length 1
($\vec{e}\cdot\vec{e}=1$)\\
$d\vec{A}$&a vector differential representing an oriented surface area
element\\ $d\vec{s}$&a vector differential representing a line element\\
$\vec{F}$& a vector representing a force\\
$H$&a scalar representing the Hamilton function of a system\\
$\tensor{I}$& the tensor for the inertia of a rigid body\\
$L$&a scalar representing the Lagrange function\\
$\vec{L}$&a vector representing an angular momentum\\
$\vec{N}$&a vector representing a torque\\
$m$&a scalar representing the mass of a particle\\
$\tensor{m}$&mass matrix in a system of coupled masses\\
$M$&the total mass of a system\\
$\vec{p}$&a vector representing the momentum $m\vec{v}$ of a particle\\
$\vec{P}$&a vector representing the total momentum of a system\\
$q_k$&a generalized coordinate\\ 
$Q$&the ``quality factor'' of a damped harmonic oscillator\\
$\vec{r}$ (or $\vec{x}$) & The position of a particle, represented by a
vector\\
$s$&path length of a trajectory\\ 
$\sigma$&scattering cross section, has unit of an area\\
$t$&time; typically a parameter that parameterizes the evolution of a system\\
$T$&either a time like a period or a total time, or a kinetic energy\\
$U$&potential energy\\
$\vec{v}$& another vector, often the velocity of a particle\\
$v$& a scalar, often referring to the speed $||\vec{v}||$\\
$W_{12}$&a scalar representing work for a state change $1\rightarrow2$\\
$\delta\vec{\theta}$& a vector representing an infinitesimal rotation by an
angle $\delta\theta$\\
$\Omega$&solid angle, characterizes a set of directions, unit sr (steradian) \\
$\vec{\omega}$& a vector representing an angular velocity\\
$x_i$&the $i$-th component of a vector $\vec{x}$\\
\end{tabular}
\clearpage

\section{Kinematics}
\subsection{Trajectories, velocity, acceleration}\label{ssec:trajectories}
This part deals with a geometric description of the trajectories of a single
point-like object without going through how such a trajectory comes about.

In its simplest form, the motion of a particle over time can be described as a
time-dependent position vector
\begin{equation}
  \vec{r}(t)=\left(\begin{array}{c}x_1(t)\\x_2(t)\\x_3(t)\end{array}\right)
  =\left(\begin{array}{c}x(t)\\y(t)\\z(t)\end{array}\right)
\end{equation}
This is a representation of the position $\vec{r}$ in traditional Cartesian
coordinates $x,y,z$ or -- to simplify notations later -- $x_1,x_2,x_3$. In
order to prepare ourselves for other coordinate systems, we write this as
\begin{equation}
\vec{r}(t)=x_1(t)\,\vec{e}_1+x_2(t)\,\vec{e}_2+x_3(t)\,\vec{e}_3=\sum_{i=1}^3x_i\vec{e}_i\,,
\end{equation}
where the $\vec{e}_i$ are the unit vectors of the standard Cartesian
coordinate system. These unit vectors are normalized, which can be expressed
by the scalar product, $\vec{e}_i\cdot \vec{e}_i=1$,
and two $\vec{e}_i$ with different indices are orthogonal,
i.e., $\vec{e}_i\cdot \vec{e}_j=0$ for $i\neq j$. This can be summarized by
the short notation
\begin{equation}
\vec{e}_i\cdot \vec{e}_j=\delta_{ij}\,,
\end{equation}
with the Kronecker delta $\delta_{ij}$ equal to 1 for $i=j$, and 0
otherwise. Such a system of unit vectors is referred to as an {\em orthonormal
  basis} for a vector, which means that each vector can be represented as a
linear combination of basis vectors, and the coordinates of any vector $\vec{x}$ can be
extracted via projection onto the corresponding unit vector,
\begin{equation}
x_i=\vec{x}\cdot\vec{e}_i\,,
\end{equation}
where the notation $(\_\cdot\_)$ denotes again the scalar product
between two vectors.

An interesting property of a trajectory of a single point that moves in time
according to $\vec{r}(t)$ is its {\em rate of change of the position}, or its
velocity. This is simply the derivation of $\vec{r}(t)$ with respect to time,
\begin{equation}
\vec{v}(t)=\lim_{\Delta t\rightarrow0}\frac{\vec{r}(t_1)-\vec{r}(t_2)}{\Delta
  t} = \frac{d\vec{r}(t)}{dt}\,,
\end{equation}
with the position at two different times $t_1, t_2$ with a time difference
$\Delta t=t_2-t_1$ and $\Delta\vec{r}=\vec{r}(t_1)-\vec{r}(t_2)$ according to
the following figure:
\begin{figure*}[h]
\includegraphics{trajectory1.eps}
\end{figure*}

The velocity as a derivative with respect to time is often written as
$\vec{v}=\dot\vec{r}$, and can be expressed as a linear combination of
coordinate base vectors with time dependent components $v_i(t)$ with $i=1,2,3$
\begin{equation}\label{eq:velocity1}
\vec{v}=\sum_{i=1}^3v_i(t)\,\vec{e}_i=\sum_{i=1}^3\dot{r}_i(t)\,\vec{e}_i\,.
\end{equation}
As can be see from the figure, the direction of the velocity vector $\vec{v}$ is
tangential to the trajectory in each point. 
Its modulus $v(t)=||\vec{v}(t)||$ is referred to as the {\em
  speed} of the point, and is obtained in the usual way via the norm of the
velocity vector,
\begin{equation}\label{eq:speed2}
v=||\vec{v}||=\sqrt{\vec{v}\cdot\vec{v}}=\sqrt{v_1^2+v_2^2+v_3^2}=\sqrt{\sum_{i=1}^3v_i^2}\,.
\end{equation} 

Apart from the parameterization of a trajectory with respect to time $t$, it is
often useful to parameterize it according to its path length $s$. This makes
it possible to describe geometrical properties of the trajectory independently
of the speed with which a point may follow it.
For example, we can express the velocity of a point via
\begin{equation}
  \vec{v}=\lim_{\Delta s\rightarrow0}\,\frac{\Delta\vec{r}}{\Delta s} \lim_{\Delta
    t\rightarrow0}\frac{\Delta s}{\Delta t} = \frac{d\vec{r}}{ds}\,\frac{ds}{dt}\,.
\end{equation}
The derivative $ds/dt$ is by definition the speed of a point, i.e., the
path length increase per unit of time. This means that we can write the
velocity vector as
\begin{equation}\label{eq:deftangential}
  \vec{v}=v\cdot\frac{d\vec{r}}{ds} =: v\cdot\vec{e}_t\,,
\end{equation}
where $\vec{e}_t=d\vec{r}/ds$ is a tangential vector $\vec{e}_t$ to the
trajectory. Contrary to $\vec{v}$, it is a unit vector, as it can be seen by
comparing 
(\ref{eq:velocity1}) and (\ref{eq:speed2}). The transition to a parameterization
of the trajectory by its path length $s$ therefore allows a definition of a 
tangential vector that only depends on the geometry of the trajectory.

We now move on to another vectorial quantity, the rate of change of velocity,
or the {\em acceleration}. This is simply the temporal derivative of
the velocity,
\begin{equation}
  \vec{a}=\frac{d\vec{v}}{dt}=\dot\vec{v}=\ddot\vec{r}
\end{equation}
Apart from just specifying a time dependence of the Cartesian coordinates
similar to (\ref{eq:velocity1}), we can express it as a linear combination of
  vectors that are associated with the trajectory. We start from (\ref{eq:deftangential}), and apply the rule for deriving
a product or two quantities:
\begin{equation}\label{eq:acceleration1}
  \vec{a}=\frac{d}{dt}(v\,\vec{e}_t)=\frac{dv}{dt}\,\vec{e}_t + v\frac{d\vec{e}_t}{dt}=\dot{v}\,\vec{e}_t + v\,\frac{d\vec{e}_t}{dt}\,.
\end{equation}
The first component in this sum is pointing in the tangential
direction, and, as its vectorial component $\vec{e}_t$ is a unit vector, the
quantity $\dot{v}$ describes the change of velocity over time along the path.
The second component in this expression contains
a temporal derivative of the tangential unit vector $\vec{e}_t$. If we take
the normalization condition for $\vec{e}_t$,
\begin{equation}
 \vec{e}_t\cdot\vec{e}_t=1\,,
\end{equation} 
and derive the whole equation with respect to time, we obtain
\begin{equation}
  \frac{d}{dt}\left(\vec{e}_t\cdot\vec{e}_t\right)=\frac{d\vec{e}_t}{dt}\cdot\vec{e}_t+\vec{e}_t\cdot\frac{d\vec{e}_t}{dt}=2\vec{e}_t\cdot\frac{d\vec{e}_t}{dt}=0\,.
\end{equation}
The vanishing dot product between product between $\vec{e}_t$ and
$d\vec{e}_t/dt$ means that the latter is a vector that is
perpendicular to the tangential vector, and thus points in a direction
orthogonal to the trajectory in each point. We now split up
$d\vec{e}_t/dt$ into a modulus and a direction, and formulate it as much as we
can by elements that only depends on the geometry  of the trajectory. For
this, we write
\begin{equation}\label{eq:curvature1}
  \frac{d\vec{e}_t}{dt}=\frac{d\vec{e}_t}{ds}\,\frac{ds}{dt}=
  v\left|\frac{d\vec{e}_t}{ds}\right|\vec{e}_n\,,
\end{equation}
where $\vec{e}_n$ is a new unit vector in the direction of the derivative of
the tangent vector $\vec{e}_t$ with respect to the path length. The last two
factors in this expression do not contain any explicit time dependence
anymore, and are thus only a property of the trajectory's geometry.

To interpret the meaning of this product, we consider a circular
trajectory parameterized by $s$,
\begin{figure*}[h]
\centerline{\includegraphics{circulartrajectory.eps}}
\end{figure*}
with
\begin{equation}\label{eq:circulartrajectory}
  \vec{r}(s)=R\left[\cos\left(\frac{s}{R}\right)\,\vec{e}_1 + \sin\left(\frac{s}{R}\right)\,\vec{e}_2\right]
\end{equation}
By taking the derivative with respect to $s$, one finds
\begin{equation}
\frac{d\vec{r}}{ds}=R\left[-\sin\left(\frac{s}{R}\right)\frac{1}{R}\,\vec{e}_1+\cos\left(\frac{s}{R}\right)\frac{1}{R}\,\vec{e_2}\right]=-\sin\frac{s}{R}\,\,\vec{e}_1+\cos\frac{s}{R}\,\,\vec{e_2}\,,
\end{equation}
making use of several derivation rules, and the fact that the unit vectors
$\vec{e}_{1,2}$ do not depend on $s$. The derivative $d\vec{r}/ds$ has the norm 1 because
of $\sin^2(s/R)+\cos^2(s/R)=1$, and is thus the unit tangential vector
$\vec{e}_t$ to the trajectory (\ref{eq:circulartrajectory}), i.e.,
$d\vec{r}/ds=\vec{e}_t$. We now proceed to
take the next derivative with respect to $s$:
\begin{equation}
\frac{d\vec{e}_t}{ds}=-\frac{1}{R}\cos\frac{s}{R}\,\vec{e}_1 - \frac{1}{R}\sin\frac{s}{R}\,\vec{e}_2=-\frac{1}{R}\left(\cos\frac{s}{R}\,\vec{e}_1+\sin\frac{s}{R}\,\vec{e}_2\right)
\end{equation}
The term in the parenthesis is again a unit vector, which we denote $\vec{e}_n$,
as it is normal to $\vec{e}_t$. Thus, the derivative 
\begin{equation}\label{eq:curvature}
  \frac{d\vec{e}_t}{ds}=-\frac{1}{R}\,\vec{e}_n
\end{equation}
for this circular trajectory has a modulus that is inversely proportional to
the radius $R$ of the circle. As any (reasonably smooth) trajectory can locally
be approximated by an arc of a circular trajectory, we now can identify the
last two terms in (\ref{eq:curvature1}) with a local radius of curvature $R$. With
this and (\ref{eq:curvature1}) we can write the expression
(\ref{eq:acceleration1}) for the acceleration as
\begin{equation}
  \vec{a}=\dot{v}\,\vec{e}_t+v^2\,\frac{1}{R}\,\vec{e}_n\,,
\end{equation}
where some care has to be taken with the direction of the unit vector
$\vec{e}_n$ of the trajectory. The quantity $1/R$ is also referred to as the
local {\em curvature} of the trajectory in any point. A straight trajectory with
$d\vec{e}_t/ds=0$ has curvature 0.

\subsection{Vectors in polar coordinates}
So far, we have chosen Cartesian coordinates to describe the trajectory of a
point, and some of its derived quantities. We now try to do this (for a
two-dimensional problem) in another coordinate system, the polar
coordinates.
\begin{figure*}[h]
  \centerline{\includegraphics{polarcoordinates.eps}}
\end{figure*}
The polar coordinates $(r,\theta)$ are connected to the Cartesian counterparts
$(x_1,x_2)$ via
\begin{equation}\label{eq:poldefinition}
  x_1=r\cos\theta\,,\quad x_2=r\sin\theta
\end{equation}
While a trajectory can simply be specified by describing the coordinate pair
$(r,\theta)$ as a function of a parameter like time $t$ or path length $s$,
the description of directed, vector-like quantities becomes more tricky. We do
need a basis to describe vectors that is somehow connected to the
coordinates. Such vectors can be constructed from the
direction specified by the position change from a given point $\vec{r}$ to
$\vec{r}+d\vec{r}$ if one of the coordinates is changed, i.e., $r\rightarrow
r+dr$ or $\theta\rightarrow\theta+d\theta$. From (\ref{eq:poldefinition}) we find (in Cartesian coordinates)
\begin{eqnarray}
  \vec{e}_r&=&const_1\cdot\frac{\partial}{\partial{r}}(x_1\,\vec{e}_1+x_2\,\vec{e}_2)=\cos\theta\,\vec{e}_1+\sin\theta\,\vec{e}_2\nonumber\\
\vec{e}_\theta&=&const_2\cdot\frac{\partial}{\partial\theta}(x_1\,\vec{e}_1+x_2\,\vec{e}_2)=-\sin\theta\,\vec{e}_1+\cos\theta\,\vec{e}_2\,,\label{eq:etheta}
\end{eqnarray}
where the constants are chosen to normalize the vectors. In the first
equation, this constant is 1, in the second equation it is $1/r$. Both vectors
$\vec{e}_r, \vec{e}_\theta$ form an orthonormal basis, that can be used to
express vectors anywhere in (here: two dimensional) space.

As an example, we now express velocity and acceleration vectors of a moving
point in these coordinates. First, we note that any point in the trajectory $\vec{r}(t)$ can be expressed as
\begin{equation}
\vec{r}=r\,\vec{e}_r
\end{equation}
We then arrive at the velocity by taking the temporal derivative:
\begin{equation}
\vec{v}=\frac{d}{dt}\left(r\,\vec{e}_r\right)=\dot{r}\,\vec{e}_r+r\,\frac{d\vec{e}_r}{dt}
\end{equation}
We use the product rule to carry out the differentiation,
because $\vec{e}_r$ will depend on the position of the point, and may thus not
be constant over time. As we are looking for changes of $\vec{e}_r$, we can
attempt to look for changes in the new coordinates:
\begin{equation}\label{eq:deriveunitvector}
  \frac{d\vec{e}_r}{dt}=\frac{\partial\vec{e}_r}{\partial\theta}\,\frac{d\theta}{dt}+\underbrace{\frac{\partial\vec{e}_r}{\partial
      r}}_{=0}\,\frac{dr}{dt}=\vec{e}_\theta\,\frac{d\theta}{dt}=\dot\theta\,\vec{e}_\theta
\end{equation}
With this, we can write the velocity vector as a linear combination of the new
unit vectors $\vec{e}_r, \vec{e}_\theta$:
\begin{equation}
  \vec{v}=\dot{r}\,\vec{e}_r+r\dot\theta\,\vec{e}_\theta
\end{equation}
Similarly, we try to express the acceleration in the basis $(\vec{e}_r,
\vec{e}_\theta)$:
\begin{equation}\label{eq:radialmotoion}
  \vec{a}=\frac{d\vec{v}}{dt}=\ddot{r}\,\vec{e}_r+\dot{r}\,\frac{d\vec{e}_r}{dt}+(\dot{r}\dot\theta+r\ddot\theta)\,\vec{e}_\theta+r\dot\theta\,\frac{d\vec{e}_\theta}{dt}
\end{equation}
We obtain the temporal derivation of $\vec{e}_\theta$ in a similar way as in
(\ref{eq:deriveunitvector}):
\begin{equation}
  \frac{d\vec{e}_\theta}{dt}=\frac{\partial\vec{e}_\theta}{\partial\theta}\,\frac{d\theta}{dt}+\underbrace{\frac{\partial\vec{e}_\theta}{\partial
      r}}_{=0}\,\frac{dr}{dt}=\dot\theta\,\frac{\partial\vec{e}_\theta}{\partial\theta}=-\dot\theta\,\vec{e}_r
\end{equation}
The latter step follows from (\ref{eq:etheta}) by differentiation of
$\vec{e}_\theta$ and comparison with $\vec{e}_r$. We now can clean up the
expression for the acceleration, and arrive at
\begin{equation}\label{eq:radialmotoion2}
  \vec{a}=(\ddot{r}-r\dot\theta^2)\,\vec{e}_r+(2\dot{r}\dot\theta+r\ddot\theta)\,\vec{e}_\theta
\end{equation}
The strategy how to obtain local unit vectors can be applied to other
coordinate systems. It is an important method to find a basis to express
vectorial quantities in whatever coordinate system is chosen.

% start of stuff from Aug 18
\subsection{Angular velocity}\label{ssec:angularvelocity1}
So far, we have encountered the vector quantities for the velocity $\vec{v}$,
which describes the rate of change of the position $\vec{r}$ of a point in
time, and the acceleration $\vec{a}$, which describes the rate of changes of
the velocity. 

We now come back to the motion of a point on a circular trajectory defined by
(\ref{eq:circulartrajectory}). Here, polar coordinates $(r,\theta)$ as
defined in (\ref{eq:poldefinition}) simplify the description because $r$ is
constant. In a trajectory is parameterized by time $t$, all information about
the motion is contained in the function $\theta(t)$. Similar to the
linear motion, we can introduce an {\em angular velocity} $\omega$ for the
instantaneous rate of change of $\theta$:
\begin{equation}
\omega(t)=\frac{d}{dt}\theta(t)
\end{equation}
This is a simple quantity for a motion in two dimensions.
% stated at the end of section \ref{ssec:trajectories}, 
As any motion in three dimensions can be locally approximated by a circular
arc, we look more carefully at a circular motion in three dimensions.
The approximate circular motion in each point has a well-defined axis of
rotation. This axis can be well described by a vector, with an ambiguity in
the pointing direction (because an axis is equally well described by
two vectors $\vec{\omega}$ and $-\vec{\omega}$), and its length. In the same
way as the rate of change of the position is encoded in the length $v=|\vec{v}|$
of the velocity vector $\vec{v}$, we now encode the rate of change of the
angle, $\omega=d\theta/dt$ of the approximate circular motion, into the length
of a vector $\vec{\omega}$, which points in the axis of rotation.  We can fix
the pointing ambiguity by postulating that $\vec{v}$, $\vec{a}=\dot\vec{v}$
(which is perpendicular to $\vec{v}$), and $\vec{\omega}$ 
form a right-handed system. Note that here the angle $\theta$ refers to a
polar coordinate system with an origin in the center of the approximating
circle. This definition of $\vec{\omega}$ is independent of any coordinate
system, a property that seems sensible to postulate for meaningful physical
quantities.

We now check the relation between $\vec{v}$, $\vec{r}$ and $\vec{\omega}$ if
we choose a coordinate system that has an origin somewhere on the axis of the
rotation as shown in the figure.
\begin{figure*}[h]
\centerline{\includegraphics{rotation2.eps}}
\end{figure*}

The modulus of the velocity is given by $v=\omega R=\omega
|\vec{r}|\sin\alpha$, and $\vec{v}$ is perpendicular both on $\vec{r}$ and
$\vec{\omega}$. This relationship is fulfilled by the vector product
\begin{equation}\label{eq:rotationspeed}
\vec{v}=\vec{\omega}\times\vec{r}\,,
\end{equation}
which incidentally also matches the choice for the direction of
$\vec{\omega}$.

In analogy to the velocity $\vec{v}$ describing the rate of change of a
position in space $\vec{r}$, and the simple relation $\vec{v}=\dot\vec{r}$,
this may suggest that there is a vector quantity that has the meaning of an
angle, and its temporal derivative leads to the vector quantity $\vec{\omega}$
above. This, however, will not work. To see why, we need to have a closer look
to coordinate transformations.

\subsection{Coordinate transformations}\label{ssec:coordinatetransformations}
It can be useful to describe a vector quantity like
the velocity in different coordinate systems. The usual Cartesian coordinate
system with some origin is one of them, the tangential coordinate system to a
trajectory, defined by $\vec{e}_t$ from (\ref{eq:deftangential}), $\vec{e}_n$
from (\ref{eq:acceleration1}) and a third unit vector,
$\vec{e}_r=\vec{e}_t\times\vec{e}_n$ orthogonal to the first two, is another
one. Yet other Cartesian reference frames can be generated from tangential
vectors to polar, spherical or cylindrical coordinate systems. They all have
the property that any vector can be expressed as a linear combination of the
respective basis vectors:
\begin{equation}
\vec{a}=\sum_{i=1}^3a_i\vec{e}_i
\end{equation}

Any
transition from a right-handed basis $\{\vec{e}_1, \vec{e}_2, \vec{e}_3\}$ to
a different right-handed basis $\{\vec{e}'_1, \vec{e}'_2, \vec{e}'_3\}$ can
be expressed as a rotation in space. This can be shown rigorously, but we skip
this here and refer to a course in basic linear algebra. The coordinate
transformation should not affect the scalar product of two vectors, and as a
consequence, norm of a single vector should be conserved, i.e,
\begin{equation}
\vec{a}'\cdot\vec{b}'=\sum_{i=1}^3{a'}_i{b'}_i=
\sum_{i=1}^3{a}_i{b}_i=\vec{a}\cdot\vec{b}\,,
\end{equation}
with components $a_i$ and $a'_i$ in the respective bases. The
transformation between the coordinates is a linear relationship, and can be
represented by
\begin{equation}\label{eq:vectortransform}
\left(\begin{array}{c}a'_1\\a'_2\\a'_3\end{array}\right)=\vec{M}\cdot
\left(\begin{array}{c}a_1\\a_2\\a_3\end{array}\right)=
\left(\begin{array}{ccc}m_{11}&m_{12}&m_{13}\\m_{21}&m_{22}&m_{23}\\m_{31}&m_{32}&m_{33}
\end{array}\right)\cdot\left(\begin{array}{c}a_1\\a_2\\a_3\end{array}\right)
\quad{\rm or}\quad
a'_i=\sum_{j=1}^3m_{ij}a_j\,,
\end{equation}
with the matrix $\vec{M}$ made up by components $m_{ij}$. The matrix
representation $\vec{M}$ of a linear transformation that preserves the scalar
product between two vectors obeys  $|{\rm det}\,\vec{M}|=1$.
%forreal-valued entries $m_{ij}$\footnote{This is a reasonable requirement,
%  because measurable physical quantities like lengths are always real-valued;
%  complex-valued components may make sense in many scenarios, but require a
%  careful analysis how to translate them back in quantities that can be
%  interpreted in a meaningful way.}. 

We now look at specific examples for rotations and their representation in the
form (\ref{eq:vectortransform}). A rotation around the $\vec{e}_3$ axis by an
angle $\phi$ is represented by
\begin{equation}\label{eq:rotation_e3}
  \vec{R}_3(\phi)=\left(\begin{array}{ccc}
      \cos\phi&-\sin\phi&0\\\sin\phi&\cos\phi&0\\0&0&1
      \end{array}\right)
\end{equation}
Matrix representations of rotations around axes $\vec{e}_{1,2}$
can be obtained by cyclic coordinate change
$1\rightarrow2\rightarrow3\rightarrow1$ (or $x\rightarrow y\rightarrow
z\rightarrow x$):
\begin{equation}
  \vec{R}_1(\phi)=\left(
    \begin{array}{ccc}
      1&0&0\\0&\cos\phi&-\sin\phi\\0&\sin\phi&\cos\phi
    \end{array}\right)
  \quad{\rm and}\quad
  \vec{R}_2(\phi)=\left(
    \begin{array}{ccc}
      \cos\phi&0&\sin\phi\\0&1&0\\-\sin\phi&0&\cos\phi
    \end{array}\right)
\end{equation}

Rotations around different axes do generally not commute. The matrix 
representing two sequential transformations is the matrix product of the
representations of the individual transformations. For example,
\begin{equation}\label{eq:noncommute1}
  \vec{R}_2(90^\circ)\cdot\vec{R}_3(90^\circ)=
  \left(\begin{array}{ccc}0&0&1\\0&1&0\\-1&0&0\end{array}\right)\cdot
  \left(\begin{array}{ccc}0&-1&0\\1&0&0\\0&0&1\end{array}\right)=
  \left(\begin{array}{ccc}0&0&1\\1&0&0\\0&1&0\end{array}\right)\,,
\end{equation}
but
\begin{equation}\label{eq:noncommute2}
  \vec{R}_3(90^\circ)\cdot\vec{R}_2(90^\circ)=
  \left(\begin{array}{ccc}0&-1&0\\0&0&1\\-1&0&0\end{array}\right)
\neq\vec{R}_2(90^\circ)\cdot\vec{R}_3(90^\circ)\,.
\end{equation}

If we were to represent a rotation by a vector, the sum of two such vectors
would be again a vector representing a rotation that should represent the
concatenation of the rotations represented by the individual vectors; this can
be 
motivated by looking at two rotations around the same axis. For different
axes, however, the representation of concatenated rotations depend on
their sequence, but the sum of two vectors does not. Hence, a rotation by a
finite angle can not be represented in a meaningful way by a vector. This
answers, in part, the problem we were facing at the end of section
\ref{ssec:angularvelocity1}: there is no underlying vector quantity which has
the angular velocity vector $\vec{\omega}$ as a rate of change in time.

We stay for a while with the properties of rotation transformations. They are
still meaningful independently of the choice of a coordinate system, although
they can not represented by vectors. Specifically, observers with different
coordinate systems can agree on a rotation axis, direction, and angle.

As rotations can be represented by matrices that transform vectors according
to (\ref{eq:vectortransform}), and such matrices can again be represented in
different coordinate systems, we can try to extract properties of rotation
transformations that are independent of the coordinate system. The determinant
${\rm det}\,\vec{M}$ of a representing matrix $\vec{M}$ is such a
property. Others are the eigenvalues of a representing matrix. For the
specific example of a rotation around $\vec{e}_3$ in 
(\ref{eq:rotation_e3}), we can evaluate the 
characteristic equation to determine the eigenvalues $\lambda$:
\begin{equation}
  |\vec{R}_3(\phi)-\lambda\vec{I}|=
  \left|
    \begin{array}{ccc}c-\lambda&-s&0\\s&c-\lambda&0\\0&0&1-\lambda
    \end{array}\right|=(1-\lambda)[(c-\lambda)^2+s^2]=0\,,
\end{equation}
with $c=\cos\phi$, and $s=\sin\phi$. This can be further simplified to 
\begin{equation}\label{eq:eigenvalue1}
  (1-\lambda)(1-2c\lambda+\lambda^2)=0\,.
\end{equation}
The first eigenvalue is $\lambda=1$; one corresponding eigenvector is
\begin{equation}
  \vec{a}=\left(\begin{array}{c}0\\0\\1\end{array}\right)=\vec{e}_3.
\end{equation}
This represents the fact that vectors along the rotation axis do not change.
More generally, an eigenvector to the eigenvalue 1 of a matrix representing
an arbitrary rotation provides a nice way to find the axis of rotation in any
Cartesian coordinate system.
For the other two eigenvalues $\lambda$ that fulfill (\ref{eq:eigenvalue1}),
we find the roots to the second factor $(1-2\cos\phi\lambda+\lambda^2)$:
\begin{eqnarray}
  \lambda&=&\cos\phi\pm\sqrt{\cos^2\phi-1}=\cos\phi\pm\sqrt{-1}\sqrt{1-\cos^2\phi}\nonumber\\
&=&\cos\phi\pm i\sin\phi=e^{\pm i\phi}
\end{eqnarray}
Therefore, we can extract the angle of rotation represented by a matrix by
looking at its complex eigenvalues. As a side remark, the corresponding
eigenvectors for a rotation represented by $\vec{R}_3(\phi)$ are
$\vec{a}=\vec{e}_1\pm i\vec{e}_2$.
These vectors are not too meaningful in classical mechanics, but become useful
e.g. when considering representations of electromagnetic fields of circular
polarization; left- and right-handed polarized fields propagating along the 
$\vec{e}_3$ axis are represented by these eigenvectors, and are preserved
under rotations around $\vec{e}_3$.

The rotations we have considered so far all preserve the handedness of a
coordinate system; they fall into the category with ${\rm
  det}\,\vec{R}=+1$; they are so-called {\em proper rotations}. Scalar-product
preserving coordinate transformations with ${\rm  det}\,\vec{R}=-1$ are
referred to as {\em improper rotations}, and are are concatenations of (an odd
number of) coordinate inversions, or mirror images with proper rotations. For
example, a reflection in direction of $\vec{e}_3$ is represented by the matrix
\begin{equation}
  \vec{M}_3=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&-1\end{array}\right)\,.
\end{equation}

We have seen examples of vectors like velocity and acceleration. For a
change to a different orthogonal coordinate system, they need to be
transformed according to (\ref{eq:vectortransform}). 
Physical quantities like temperature in a given location or
the speed of a point that do not change under coordinate changes form the
class of {\em scalars}. We can extend
this concept and ask e.g. how to transform the representations of physical
properties that need to be represented by matrices. Such quantities, apart
from the rotations we just saw, do exist; an example in continuum mechanics
would be the {\em stress tensor}), but this is beyond the scope 
of this module. The moment of inertia of a solid is another one. 
If such a quantity (referred to as {\em tensor of rank 2}) is 
represented by a matrix $\vec{T}$ made up by components $t_{ij}$, its
components transform under $\vec{M}$ according to 
\begin{equation}\label{eq:tensortransform}
{t'}_{ij}=\sum_{k,l=1}^3m_{ik}m_{jl}t_{kl}
\end{equation}

One can turn these transformation properties around, and classify a physical
property as scalar, vector or tensor according to their 
transformation properties: Scalars do not change under rotations, vector are
transformed according to (\ref{eq:vectortransform}), tensors of rank 2
according to (\ref{eq:tensortransform}), and so on.

\subsection{Infinitesimal rotations}\label{ssec:smallrotations}
Examples (\ref{eq:noncommute1}) and (\ref{eq:noncommute2}) show that finite
rotations generally do not commute. However, the situation changes if we
consider only small rotations. Starting from
(\ref{eq:rotationspeed}), we find a change of a position $\vec{r}$
in a small time interval $\delta t$:
\begin{equation}\label{eq:smallmove}
\vec{r}\rightarrow\vec{r}+\delta \vec{r}_1\quad{\rm with}\quad
\delta \vec{r}_1=\vec{v}\delta t = (\vec{\omega}\delta t) \times \vec{r}=
\delta\vec{\theta}_1 \times \vec{r}\,,
\end{equation}
where $\delta\vec{\theta}_1$ represents a small rotation by an angle
$\delta\theta_1$ around an axis parallel to $\delta\vec{\theta}_1$, and a
handedness in the same way as defined for $\vec{\omega}$.

Two sequential infinitesimal rotations, $\delta\vec{\theta}_1$ followed by
$\delta\vec{\theta}_2$, change the position $\vec{r}$ of a point similarly to
(\ref{eq:smallmove}) by the infinitesimal amount 
\begin{eqnarray}
  \delta\vec{r}_{12}&=&\delta\vec{r}_1+ 
  \delta\vec{\theta}_2\times(\vec{r}+\delta\vec{r}_1)\nonumber\\
  &=&\delta\vec{\theta}_1\times\vec{r}+
  \delta\vec{\theta}_2\times(\vec{r}+\delta\vec{\theta}_1\times\vec{r})\nonumber\\
&=&\delta\vec{\theta}_1\times\vec{r}+\delta\vec{\theta}_2\times\vec{r}+
\delta\vec{\theta}_2\times\delta\vec{\theta}_1\times\vec{r}\nonumber \\
  &\approx&\delta\vec{\theta}_1\times\vec{r}+\delta\vec{\theta}_2\times\vec{r}
\end{eqnarray}
by neglecting the product term of two infinitesimal quantities in the last
step. This representation is the same if one carries out the rotations in
different order. As a consequence, infinitesimal rotations represented by
vectors $\delta\vec{\theta}$ {\em do commute}.

We can also see this with an infinitesimal version of our example in
(\ref{eq:noncommute1}) and (\ref{eq:noncommute2}). We first approximate
the representation (\ref{eq:rotation_e3}) by a truncated Taylor expansion for
small angles $\delta\theta$: 
\begin{equation}
  \vec{R}_3(\delta\theta_3)=\left(
    \begin{array}{ccc}
      1-\delta\theta_3^2/2+\ldots&-\delta\theta_3+\delta\theta_3^3/3!-\ldots&0\\
      \delta\theta_3-\delta\theta_3^3/3!+\ldots&1-\delta\theta_3^2/2+\ldots&0\\
      0&0&1
    \end{array}\right)
  \approx\left(
    \begin{array}{ccc}
      1&-\delta\theta_3&0\\\delta\theta_3&1&0\\0&0&1
    \end{array}\right)\,,
\end{equation}
again by neglecting terms in higher than linear order in $\delta\theta$.
Similarly, we approximate 
$\vec{R}_2(\delta\theta_2)$, and evaluate two concatenated
small rotation matrices. We find
\begin{equation}
  \vec{R}_2(\delta\theta_2)\cdot\vec{R}_3(\delta\theta_3)=
  \left(
    \begin{array}{ccc}
     1&-\delta\theta_3&\delta\theta_2\\\delta\theta_3&1&0\\-\delta\theta_2&\underline{\delta\theta_2\delta\theta_3}&1
   \end{array}\right)
\end{equation}
and
\begin{equation}
  \vec{R}_3(\delta\theta_2)\cdot\vec{R}_3(\delta\theta_3)=
  \left(
    \begin{array}{ccc}
     1&-\delta\theta_3&\delta\theta_2\\\delta_3&1&\underline{\delta\theta_2\delta\theta_3}\\-\delta\theta_2&0&1
   \end{array}\right)\,.
\end{equation}
The two expressions differ only by the two underlined terms, which is a
product of two infinitesimal angles. Neglecting them with the same argument
used for truncating the Taylor expansion, the infinitesimal rotations commute.

So in summary, an infinitesimal rotation can be represented by a vector
$\delta\vec{\theta}$. The direction of this vector characterizes the rotation
axis, and can be transformed in a meaningful way according to
(\ref{eq:vectortransform}).
Its modulus represents the rotation angle, and the sum of two vectors
$\delta\vec{\theta}_1+\delta\vec{\theta}_2$ represents correctly the
concatenation of the two individual infinitesimal rotations. 

\section{Newtonian Mechanics for single particles}
So far, we only looked at a mathematical description of trajectories. One
of the main goals in classical mechanics is to describe the evolution of a
physical system in time. An extremely successful method for this was developed
by Sir Isaac Newton. The idea is not to specify the whole trajectory of a
particle, but to combine very few underlying universal principles with
standard mathematical methods. These principles were expressed as ``laws'', 
and are sufficient (together with some initial parameters) to lead to a
complete description of the evolution of a physical system. 

These laws, together with the mathematical tools to solve differential
equations, have turned out to be tremendously successful in the description of
mechanical systems over an extremely wide scales in space and time, ranging
from motion on the molecular level up to the motion of planets. It was not
until perhaps a 120 years ago that the descriptive strength of these few
``laws'' has been seen as incomplete for describing the motion of mechanical
objects, specifically for areas where very small masses, extremely short time
scales, high energies, and velocities on the order of the speed of light are
involved.

That said, Newton's laws turned physics into a science with an enormous
predictive power based on very few and simple rules, and cover 
extremely well the phenomena that we encounter in our daily life.

\subsection{Newton's laws}
Sir Isaac Newton formulated these laws in 1687 in his ``Principia''. They were
formulated as follows:
\begin{itemize}
\item[I.] {\em A body remains at rest or in uniform motion unless acted upon a
    force.}
\item[II.] {\em A body acted upon a force moves in such a manner that the time
  rate of change of momentum equals the force.}
\item[III.] {\em If two bodies exert forces on each other, these forces are
    equal in magnitude and opposite in direction.}
\end{itemize}


These laws do not mean that the physical
world has to follow them strictly --- they are more a tool to
efficiently describe how many things evolve in time. In a sufficiently well
specified context of initial conditions and participating forces, they 
and allow a very accurate description of the motion of objects.
However, it is necessary to observe how well Newton's laws (or any other
physical law) capture a phenomenon in nature to see if corrections or 
additional principles need to be added.

When these laws were formulated, the whole mathematical language of
definitions and formal logic was not as formally developed as it is now. Thus,
it makes sense to comment on these laws to make clearer what they actually
state.

The first law is basically a definition of a ``free particle''. The notion of
a force is used, but not introduced in a too useful way.
``Uniform motion'' means that the velocity of a particle does not change in
time. For making a statement about a velocity, we do require a specific
reference frame, i.e., the choice of an origin and a few other properties
that we will see later.  Formally, we would write the first law as
\begin{equation}\label{eq:FirstLaw}
\vec{F}=0 \Rightarrow\vec{v}=const.
\end{equation}

The second law explicitly relates force with momentum (called {\em
  quantity of motion} originally), and Newton provided a definition of  the
momentum of a particle as
\begin{equation}\label{eq:momentumdefinition}
\vec{p}=m\vec{v}\,,
\end{equation}
where $m$ is the mass of a particle, and $\vec{v}$ its
velocity. Mathematically, the second law can be written as
\begin{equation}\label{eq:ForceDefinition}
\vec{F}=\frac{d}{dt}\vec{p}=\frac{d}{dt}(m\vec{v})=m\vec{a}=m\ddot\vec{x}\,.
\end{equation}
This is a definition of what is meant by {\em force}. However, this definition
requires that one has already an idea what the mass $m$ of an object
is. While it seems intuitively clear, this is a reference to the {\em inertial
  mass} of an object. It is an intrinsic quantity of an object that is not
subject of its state of motion.

The third law is a statement on the motion of {\em two objects} that exert
forces on each other. Even more specific, it makes reference to forces aligned
along the connecting line between them, 
\begin{equation}
\vec{F}_{12} \parallel (\vec{x}_1-\vec{x}_2)\,,
\end{equation}
where $\vec{x}_1$ and $\vec{x}_2$ are the positions of the two objects.
Such forces are referred to as {\em central forces}, but the choice of the
name ``central'' will only become clear at a stage when we consider objects of
finite size. Examples of such forces are forces exerted by a elastic spring
connecting the two bodies, the gravitational force between two heavy masses,
or the force between two electrically charged objects, van der Waals forces,
and others.
\begin{figure*}[h]
\centerline{\includegraphics{newton3.eps}}
\end{figure*}

In such a case, the third law then states that
\begin{equation}\label{eq:thirdlaw}
\vec{F}_{12}=-\vec{F}_{21}
\end{equation}

Notably, the third law does not apply to forces that depend on the velocity of
particles. Examples for such forces are the friction of an object moving
through a medium, or even the (weak) velocity dependence of the gravitational
force.

Together with the definition of force in the second law in
(\ref{eq:ForceDefinition}), one can write the third law in its version
(\ref{eq:thirdlaw}):
\begin{equation}
m_1\frac{d\vec{v}_1}{dt} = - m_2\frac{d\vec{v}_2}{dt}\quad{\rm or}\quad
m_1\vec{a}_1 = -m_2\vec{a}_2\quad{\rm or}\quad
\frac{m_1}{m_2}=-\frac{a_2}{a_1}\,,
\end{equation}
where the minus sign indicates the opposite orientation of the two
accelerations on the two masses. This relation can be used to compare
the two inertial masses by comparing the accelerations of the two bodies when
they exert a force on each other. Again, an appropriate
reference frame is necessary to determine the accelerations. 

We often do not measure the ratio of inertial masses, but that of {\em heavy
  masses} in the gravitational field of the earth with a balance. It does not
follow by logic from Newton's law that the heavy mass and 
the inertial mass used in the definition of a force (\ref{eq:ForceDefinition})
are the same. However, many tests have been carried out that compared heavy and
inertial masses that seem to indicate that they are indeed the same
property. Often Galileo is attributed to have carried out the first of these
tests by comparing the falling time of balls of the same material, but
different masses from the tower of Pisa some time around 1600, but apparently 
Simon Stevin seemed to have actually carried out this experiment on the church
tower in Delft in 1586. Those early experiments had a limited accuracy, and
experiments carried out by Newton himself seemed to have shown the equivalence
of heavy and inertial mass ``only'' to within $10^{-3}$. Much more accurate
tests were carried out by E\"otv\"os\footnote{\href{https://en.wikipedia.org/wiki/Lor\%C3\%A1nd_E\%C3\%B6tv\%C3\%B6s}{Lor\'and E\"otv\"os}, 1848-1919} in 1890, and \href{https://dx.doi.org/10.1016/0003-4916(64)90259-3}{more recent experiments} by
Dicke\footnote{\href{https://en.wikipedia.org/wiki/Robert_H._Dicke}{Robert H. Dicke}, 1916-1997} in 1964 could show the equivalence of the two quantities to within
$10^{-12}$. This seems to suggest that they are indeed equivalent. The
assertion that the two are equivalent is referred to as the {\em equivalence
  principle}, and is one of the cornerstones of general theory of relativity.

Another important consequence of the third law can be seen when using the
definition of the force in the form of
$\vec{F}=d\vec{p}/dt$ in an isolated system of two bodies 1,2. Then, it states
\begin{equation}\label{eq:conservmomentum1}
\frac{d\vec{p}_1}{dt}=-\frac{d\vec{p}_2}{dt}\quad{\rm or}\quad
\frac{d}{dt}(\vec{p}_1+\vec{p}_2)=0\,.
\end{equation}
This means that the sum $\vec{p}_1+\vec{p}_2$ of the two momenta is
constant in time, i.e., the total momentum in this system is {\em
  conserved}. We will see a few more of such conservation laws, they tend to
simplify the description of the dynamics of a system.

\subsection{Inertial reference frames}
The consequences of Newton's laws only make sense if there is an appropriate
coordinate system that is used to describe the derivatives of the momentum via
(\ref{eq:thirdlaw}). Such coordinate systems are referred to as {\em inertial
  reference frames}. 
Therefore, a somewhat lame definition of such a reference system is can be
seen as one in which Newton's laws hold. 
Then, the first law (\ref{eq:FirstLaw}) can be used as a
definition: An {\em inertial reference frame} is a reference frame in which
a body under no influence of a force remains in uniform motion.

Once one has identified an inertial reference frame, there are several
others: As Newton's laws make statements about the {\em change} of velocities
of bodies, the addition of a constant velocity does not affect the validity
of the laws. Thus, a coordinate transformation between two reference systems
$K$ and $K'$ with a position vector of a point $\vec{r}$ in $K$, and
$\vec{r}'$ in $K'$ according to
\begin{equation}
\vec{r}\rightarrow\vec{r}'=\vec{r}+\vec{v}_0t
\end{equation}
with a constant velocity $\vec{v}_0$ leads to
$\ddot\vec{r}=\ddot\vec{r}'$. Thus, if Newton's laws hold in $K$, they also do
in $K'$. It should be noted that the times $t$ and $t'$ in the two
reference systems are assumed to be the same. Such transformations are
referred to as {\em Galilean transformations}, the equivalence of two inertial
frames under such a transformation is referred to as {\em Galilean
  invariance}, or the principle of {\em Newtonian or Galilean relativity}.

\subsection{Equation of motion}
We can now use Newton's second law (\ref{eq:ForceDefinition}) to determine the
dynamic of a system, simply by considering 
\begin{equation}
\vec{F}=m\ddot\vec{x}
\end{equation}
as a differential equation for the position $\vec{x}$ of a body. This is a
second order {\em ordinary differential equation} (ODE), which we can solve if
we know the force $\vec{F}$ acting on the body, and sufficient initial or
boundary conditions for the position and/or velocities. However, this equation
of motion does not always hold. Examples where this is {\em not} applicable are
\begin{itemize}
\item situations in non-inertial reference frames, like rotating
  reference frames. This becomes significant e.g. in
  the description of the motion of air particles and clouds on long ranges;
\item at velocities that are not small compared to the speed of light;
\item when time would be not homogenous. This is perhaps a rather exotic
  restriction.
\end{itemize} 

\subsubsection{Free Fall}
A simple equation of motion is presented for a constant force
$\vec{F}=\vec{F}_0$. Then, the acceleration $\vec{a}=\vec{F}_0/m$ is constant
as well, and we can obtain the velocity by integration over time:
\begin{equation}\label{eq:eqmot1}
\vec{v}(t)=\int\limits_0^{t}\vec{a}(t')dt'+\vec{v}_0 =
\frac{1}{m}\vec{F}_0\,t+\vec{v}_0\,,
\end{equation}
where $\vec{v}_0$ is the velocity of the particle at time $t=0$. The
position $\vec{x}(t)$ of the particle at any time $t$ follows from integration
of $\vec{v}(t)$,
\begin{equation}\label{eq:eqmot2}
\vec{x}(t)=\int\limits_0^t\vec{v}(t')dt'+\vec{x}_0 =
\frac{1}{2m}\vec{F}_0t^2+\vec{v}_0t+\vec{x}_0\,,
\end{equation}
With the additional integration constant $\vec{x}_0$ fixing the position at
$t=0$.
A typical example for such a case is the acceleration of a body with a mass
$m$ under the influence of gravity on earth, ignoring eventual friction
forces. Such a motion is called {\em free fall}, but can have initial
conditions that include a velocity component that points in an arbitrary
direction.
\begin{figure*}
\centerline{\includegraphics[width=0.85\textwidth]{freefall1.eps}}
\end{figure*}

This is a nice example where one can see that the vectorial equation of motion
(and its solution) separates into individual components. In the example shown
in the figure,
the constant force $\vec{F}_0$ has no component in $x$ direction. Therefore, the
motion in $x$ is uniform, i.e., the velocity component $v_x$ is constant, and
the component $x(t)$ increases linearly in time. For the component in $y$, the
motion is quadratic in $t$. The choice of a suitable coordinate system
therefore can simplify the integration of the equation of motion, and separate
out certain degrees of freedom.

\subsubsection{Friction Forces}\label{ssec:friction}
Another class of forces are friction forces. They appear when an object moves
in a fluid, or in contact with a surface, and opposes the motion
of a particle. They depend on the velocity either linearly,
\begin{equation}\label{eq:friction1}
\vec{F}=-\alpha\vec{v}
\end{equation}
with a positive constant $\alpha$, or more generally,
\begin{equation}\label{eq:friction2}
\vec{F}=-f(v)\frac{\vec{v}}{v}
\end{equation}
with a function $f(v)$ depending on the speed $v$ of the object. The
dependency can be rather complex and depends on the nature of the friction,
and on the velocity range. If something moves through a fluid
like air slow enough, $f(v)\propto v$ as in (\ref{eq:friction1}), but for
higher speeds, a quadratic dependency is observed. An approximate heuristic
formulation of such a friction force that is used to describe the drag of
objects like cars or bicycles moving through air is
\begin{equation}\label{eq:friction3}
f(v)=\frac{1}{2}c_W\rho A v^2\,,
\end{equation}
with the density of air $\rho$, the surface $A$ area of the vehicle, and a
dimensionless shape-dependent drag coefficient $c_W$ that is around 0.3 for
most cars these days, and around 1 for someone on a bicycle for typical speed
ranges. If $v$ comes close to the speed of sound, $f(v)$ undergoes a
relatively complex change.

% \section{Methods of solving equations of motion}
% The solving of an equation of motion originating from typical mechanical
% problems is really a mathematical discipline. However, a few more
% examples are so common that they deserve a special mentioning.

\subsubsection{Harmonic oscillator}\label{sssec:harmonicosc1}
The motion of a harmonic oscillator is a particularly important and
appears in many areas of physics. We consider the simple case where a mass $m$
is connected to a spring with a Hook constant $k$, and motion only takes place
in $x$ direction:
\begin{figure*}[h]
\centerline{\includegraphics{harmonicoscillator1.eps}}
\end{figure*}

The spring exerts a restoring force $F_x$ on the mass,
\begin{equation}\label{eq:harmosc1}
F_x=-kx\,,
\end{equation}
and with (\ref{eq:ForceDefinition}) or $F_x=m\ddot{x}=-kx$, the equation of
motion takes the form 
\begin{equation}\label{eq:harmosc2}
\ddot{x}(t)+\frac{k}{m}x(t)=0\,.
\end{equation}
This is a linear ordinary differential equation, because all derivatives of $x$
appear linearly. Such equations have special solutions
\begin{equation}\label{eq:lineq:sol1}
x(t)=e^{st}
\end{equation}
with a (generally complex) constant $s$. Inserting (\ref{eq:lineq:sol1}) into
(\ref{eq:harmosc2}) leads to 
\begin{equation}
\left(s^2+\frac{k}{m}\right) x(t)=0\,.
\end{equation}
As this equation has to hold for all times, the first term needs to vanish:
\begin{equation}\label{eq:chareq}
s^2+\frac{k}{m}=0
\end{equation}
This is  referred to as the characteristic equation of
(\ref{eq:harmosc2}). Its two roots are
\begin{equation}\label{eq:harmoscfreq}
s_{1,2}=\pm\sqrt{-\frac{k}{m}}=\pm i\sqrt{\frac{k}{m}}=\pm i\omega
\quad{\rm with}\quad\omega=\sqrt{\frac{k}{m}}
\end{equation}
Because the roots $s_{1,2}$ are distinct, the general solution of
(\ref{eq:harmosc2}) is a  linear combination of the two solutions
(\ref{eq:lineq:sol1}) for $s_1$ and $s_2$,
\begin{equation}\label{eq:lineq:sol2}
x(t)=a_1e^{s_1t}+a_2e^{s_2t}=a_1e^{i\omega t}+a_2e^{-i\omega t}\,.
\end{equation}
The constants $a_{1,2}$ are now chosen to meet the initial conditions of
the differential equation. To fully determine the motion in time, exactly two
initial conditions are needed; let's
assume $x(t=0)=x_0$, and $v(t=0)=\dot{x}(t=0)=0$. Inserting these
conditions into (\ref{eq:lineq:sol2}) leads to
\begin{eqnarray}\label{eq:lineq:sol3}
x(t=0)&=&a_1+a_2=x_0\\
\dot{x}(t=0)&=&a_1i\omega-a_2i\omega=i\omega(a_1-a_2)=0
\end{eqnarray}
The second part leads to $a_1=a_2$, and the first part then to $a_1=x_0/2$, so
we finally get the oscillatory solution to (\ref{eq:harmosc2}) that meets the
initial conditions,
\begin{equation}\label{eq:lineq:sol4}
x(t)=\frac{x_0}{2}\left(e^{i\omega t}+e^{-i\omega t}\right)=x_0\cos\omega t\,,
\end{equation}
with the oscillatory solution of amplitude $x_0$ and a period $2\pi/\omega$:
\begin{figure*}[h]
\centerline{\includegraphics{harmonicoscillator2.eps}}
\end{figure*}

\subsection{Angular motion}
Newton's laws introduce the concept of momentum, which is changed by a force,
and can be used to make a statement about the linear motion of a body. If no
force is acting on the body, the linear momentum $\vec{p}$ is
conserved. Similarly for angular motion, one can formally define a vector
termed {\em angular momentum} $\vec{L}$:
\begin{equation}\label{eq:angularmomentumdef}
\vec{L}:=\vec{r}\times\vec{p}\,,
\end{equation}
where $\vec{r}$ is the position of a particle with respect to some origin $O$,
and $\vec{p}=m\vec{v}$ its linear momentum. This definition does not imply any
circular motion, but it is inspired by it, as shown below:
\begin{figure*}
\centerline{
  \includegraphics{rotation3.eps}
  \includegraphics{torque.eps}
}
\end{figure*}
In the case of a circular trajectory and a constant speed $v$, the newly
defined vector $\vec{L}$ is constant, and points in the direction
of the earlier introduced angular velocity $\vec{\omega}$.

The temporal derivative of the angular momentum  defined in
(\ref{eq:angularmomentumdef}) is given by:
\begin{eqnarray}
\dot\vec{L}&=&\frac{d}{dt}\left(\vec{r}\times\vec{p}\right)\nonumber\\
&=&\dot\vec{r}\times\vec{p}+\vec{r}\times\dot\vec{p}\nonumber\\
&=&\underbrace{\vec{v}\times(m\vec{v})}_{=0}+\vec{r}\times\dot\vec{p}\nonumber\\
&=&\vec{r}\times\vec{F}\,,\\
\end{eqnarray}
where $\vec{F}$ is the force acting on the particle. Similar to the definition
of $\vec{L}$ in (\ref{eq:angularmomentumdef}), one can define a vector
$\vec{N}$ called {\em torque} with respect to an origin $O$:
\begin{equation}\label{eq:torquedefinition}
\vec{N}:=\vec{r}\times\vec{F}
\end{equation}
With this definition, the temporal derivative of $\vec{L}$ is simply given by
\begin{equation}\label{eq:angularmomemtumchange}
  \dot\vec{L}=\vec{N}
\end{equation}
This looks similar to the second law for linear momentum,
$\dot\vec{p}=\vec{F}$, but for rotations. 
Thus, for angular motion, angular momentum $\vec{L}$ and torque $\vec{N}$
play similar roles as momentum $\vec{p}$ and force $\vec{F}$ for linear
motion.

If $\vec{N}=0$, $\dot\vec{L}=0$, i.e., the angular momentum $\vec{L}$ remains
constant or {\em  is conserved}. 
An important example for this situation can be found with central force
problems, like the Coulomb interaction or gravitational attraction between two
heavy bodies. If the origin $O$ for determining the angular momentum is chosen
in one of the two bodies, or somewhere on the connecting line between them, the
force $\vec{F}$ exerted by each body to the other one is parallel to
$\vec{r}$. Then, 
\begin{equation}
\vec{N}=\vec{r}\times\vec{F}=\vec{r}\times(const\cdot\vec{r})=0\,.
\end{equation}
Consequently, the angular momentum does not change in time: isolated systems
governed by a central force (planets around a central star, electrons 
around a proton) conserve the angular momentum. This does not mean,
however, that the trajectories have to be circular!

\section{Work and Energy}
While Newton's laws allow for a straightforward formulation of equations of
motion via forces, they require a knowledge of the vectorial quantity
force. This can become a bit complicated, especially 
if forces need to be introduced to fulfill boundary conditions and
constraints. Sometimes a quantity of interest may not require to fully solve
the equations of motion. On the other hand, some of the expressions for forces
can get simpler as well with the introduction of concepts of work and
energy. They will also be required for deriving the equations of motion of a
physical system in different ways than using Newton's laws.

\subsection{Kinetic energy}\label{ssec:kineticenergy}
We start with the definition of an integral quantity between two positions 1
and 2 of a particle along a path $\Gamma$ as indicated:

\begin{equation}\label{eq:workdefinition}
 W_{12}:=\int\limits_{\Gamma_{1\rightarrow2}}\vec{F}\cdot d\vec{s}\qquad
  \raisebox{-1.2cm}{\includegraphics{pathintegral1.eps}}
\end{equation}
This is definition of ``work'' $W_{12}$ for a transition between two points
integrates up the scalar products between a local force $\vec{F}$ and line
elements $d\vec{s}$ along the path $\Gamma$. It can be evaluated component-wise,
\begin{equation}
\int_\Gamma\vec{F}\cdot d\vec{s}=\int_\Gamma\sum\limits_iF_i dx_i
\end{equation}
Often one can also evaluate the kernel $\vec{F}d\vec{s}$ of this integral with
respect to a sensible parameterization of the path $\Gamma$, like the path
length $s$ introduced earlier,
\begin{equation}
\vec{F}\cdot d\vec{s}=\vec{F}\cdot\vec{e}_t\,ds\,,
\end{equation}
where $\vec{e}_t$ is the tangential vector to the path, and $ds$ a length
element. 
With $\vec{F}=m\dot\vec{v}$,
and parameterizing via time, $d\vec{s}=\vec{v}dt$, it becomes
\begin{eqnarray}\label{eq:kineticenergydifferential}
\vec{F}\cdot d\vec{s}&=&(m\dot\vec{v})\cdot(\vec{v}\,dt)\nonumber\\
&=&\frac{1}{2}m\,(2\vec{v}\dot\vec{v}\,dt)\nonumber\\
&=&\frac{1}{2}m\,\,d\left(\vec{v}\cdot\vec{v}\right)=d\left(\frac{1}{2}m\,\vec{v}^2\right)
\end{eqnarray}
The third step in the development above can be seen as reversing the
differentiation of the scalar product $\vec{v}\cdot\vec{v}=:\vec{v}^2$ with
respect to time:
\begin{equation}
\frac{d}{dt}\left(\vec{v}\cdot\vec{v}\right)=\vec{v}\cdot\dot\vec{v}+\dot\vec{v}\cdot\vec{v}=2\vec{v}\dot\vec{v}
\end{equation}
The last step in (\ref{eq:kineticenergydifferential}) is just combining the
constants into a single differential $d(m\vec{v}^2/2)$. By defining a
quantity
\begin{equation}\label{eq:kineticenergydefinition}
T:=\frac{1}{2}m\vec{v}^2\,,
\end{equation}
one can simplify the differential $d(m\vec{v}^2/2)=dT$, and the work
integral becomes
\begin{equation}\label{eq:workandT}
W_{12}=\int\limits_{\Gamma_{1\rightarrow2}}\vec{F}\cdot d\vec{s}
=\int\limits_{\Gamma_{1\rightarrow2}}dT=T_2-T_1
\end{equation}
This is a somewhat tricky step: by showing that $\vec{F}\cdot d\vec{s}$ can be
written as a full differential $dT$ in the newly defined quantity $T$, the
integration becomes trivial, and the result is the difference between the
two values of $T$ at points 1 and 2. 

The quantity $T$ only depends on the velocity $\vec{v}$ in a given point; it
is therefore reasonable to call the definition
(\ref{eq:kineticenergydefinition}) {\em kinetic energy} $T$ of a
particle. According to (\ref{eq:workandT}), the work $W_{12}$ along a path
$\Gamma$ evaluates the change in the kinetic energy of the particle moving
from point 1 to 2.

This statement holds independently of the form of the force, which
can depend on the position and velocity of the particle, and explicitly
on time: $\vec{F}=\vec{F}(\vec{r},\vec{v},t)$.

Also note that definition (\ref{eq:kineticenergydefinition}) is
{\em reasonable} in the sense that $T=0$ for $\vec{v}=0$, but
not {\em unique}: we could have chosen to add a constant
$T_0$ to $T$. Such an additive
constant leaves both the full differential $dT$ unchanged, and cancels
out in the subtraction (\ref{eq:workandT}). This is something to pay attention
to whenever some quantity (here: $T$) is defined that
conveniently describes differential property (here: $dT=\vec{F}\cdot d\vec{s}$).

\subsection{Conservative forces and potential energy}\label{ssec:conservativeforce}
For forces that only depend on the position $\vec{r}$ of a particle, one can
assign a ``field'' $\vec{F}(\vec{r})$. The work integral
(\ref{eq:workdefinition})
\begin{equation}
W_{12}=\int\limits_{\Gamma_{1\rightarrow2}}\vec{F}(\vec{r})\cdot d\vec{s}
\end{equation}
can be more directly evaluated with results from vector calculus. In some
cases, the work does not even depend 
on the specific path $\Gamma$ a particle takes, but only on the end points 1
and 2.  
In such cases,  $\vec{F}(\vec{r})$ is called a  {\em
  conservative} force.
\begin{figure*}[h]
\centerline{\includegraphics{pathintegral2.eps}}
\end{figure*}

A path integral between points 1 to 2 along a path $\Gamma$ can be split up
in an integral between points 1 and an intermediate point 3, and one between
points 3 and 2:
\begin{equation}
\raisebox{-4mm}{\includegraphics{pathintegral3.eps}}\quad
\int\limits_{\Gamma_{1\rightarrow2}}\!\!\vec{F}\cdot d\vec{s}\,=\!\!
\int\limits_{\Gamma_{1\rightarrow3}}\!\!\vec{F}\cdot d\vec{s}\,+\!\!
\int\limits_{\Gamma_{3\rightarrow2}}\!\!\vec{F}\cdot d\vec{s}
\end{equation}
If the path integral is independent of the path $\Gamma$, and therefore any
intermediate point 3 like in the path above, it has to take the
form of a difference of an endpoint-dependent function:
\begin{equation}\label{eq:potentialdifference}
\int\limits_{\Gamma_{1\rightarrow2}}\!\!\vec{F}\cdot d\vec{s}\,=U(\vec{r}_1)-U(\vec{r}_2)
\end{equation}
This form suggests that the integral kernel can be written as a
full differential, $\vec{F}\cdot d\vec{s}=-dU$. With the ansatz
\begin{equation}\label{eq:gradientforce}
\vec{F}=-\nabla U=\sum\limits_i\frac{dU}{dx_i}\vec{e}_i\,,
\end{equation}
one can verify that the path integral takes indeed the form
(\ref{eq:potentialdifference}):
\begin{eqnarray}
\int\limits_{\Gamma_{1\rightarrow2}}\!\!\vec{F}\cdot d\vec{s}&=&
-\int\limits_{\Gamma_{1\rightarrow2}}\!\!(\nabla U)\cdot d\vec{s}\nonumber\\
&=&-\int\limits_{\Gamma_{1\rightarrow2}}\!\!\left(\sum\limits_i\frac{\partial U}{\partial
    x_i}\vec{e}_i\right)\left(\sum\limits_j\vec{e}_jdx_j\right)\nonumber\\
&=&-\int\limits_{\Gamma_{1\rightarrow2}}\!\!\sum\limits_i\frac{\partial
  U}{\partial x_i}dx_i\,=\,
-\!\!\int\limits_{\Gamma_{1\rightarrow2}}\!\!dU\nonumber\\
&=& U(\vec{r}_1)-U(\vec{r}_2)
\end{eqnarray}

Similarly to the definition of the kinetic
energy, the function $U(\vec{r})$ is not uniquely defined. By adding a
constant $U_0$ to a given function $U(\vec{r})$, the resulting force and
therefore also the work integral $W_{12}$ do not change, because $U$ only
appears as a derivative. The scalar function $U$ has the same dimension as the 
previously defined kinetic energy, hence, it seems reasonable to refer to this
quantity as {\em potential energy} or simply {\em potential\/}.

To see better when a force field can be written as the gradient of a potential
$U(\vec{r})$, we consider two paths $\Gamma_A$ and $\Gamma_B$ from point 1 to
2. By definition, the work integral $W_{12}$ for a conservative field will be
the same for both paths:
\begin{equation}
\raisebox{-5mm}{\includegraphics{pathintegral3a.eps}}\qquad
\int\limits_{\Gamma_{A, 1\rightarrow2}}\!\!\vec{F}\cdot d\vec{s}=
\int\limits_{\Gamma_{B, 1\rightarrow2}}\!\!\vec{F}\cdot d\vec{s}
\end{equation}
Reversing one of the two paths changes the sign of the work integral:
\begin{equation}
W_{21}=\int\limits_{\Gamma_{-B, 2\rightarrow1}}\!\!\vec{F}\cdot d\vec{s}=
-\int\limits_{\Gamma_{B, 1\rightarrow2}}\!\!\vec{F}\cdot d\vec{s}=-W_{12}
\end{equation}
The concatenated path
$1\stackrel{\Gamma_A}{\longrightarrow}2\stackrel{\Gamma_{-B}}{\longrightarrow}1$ is a closed, and
the path integral vanishes:
\begin{equation}\label{eq:nowork}
\raisebox{-5mm}{\includegraphics{pathintegral4.eps}}\qquad
\oint\limits_{\Gamma_{A}+\Gamma_{-B}}\!\!\vec{F}\cdot
d\vec{s}=W_{12}+W_{21}=0
\end{equation}
The circle over the integral symbol is a convention indicating the closed path
integration. One of the results of vector calculus, referred to as
\href{https://en.wikipedia.org/wiki/Stokes'_theorem}{Stokes' theorem}, relates the path integral of a vector field $\vec{F}$ along the
boundary $\partial S$ of an orientable surface $S$ with a surface integral of
the curl $\nabla\times\vec{F}$ of the field:
\begin{equation}\label{eq:stokes}
\raisebox{-1.5cm}{\includegraphics[width=5cm]{stokes.eps}}\qquad
\oint\limits_{\partial{S}}\vec{F}\cdot d\vec{s}=
\int\limits_S(\nabla\times\vec{F})\cdot d\vec{A}
\end{equation}
As closed path integrals of the type (\ref{eq:nowork}) vanish for conservative
fields for all paths, the integral over $\nabla\times\vec{F}$ on the right
side is also identical to zero. This must hold for all surfaces, also
sufficiently small ones where $\nabla\times\vec{F}$ is smooth, which implies
that 
\begin{equation}
  \nabla\times\vec{F}=0
\end{equation}
everywhere. This is an important result: the curl of a conservative force
field vanishes -- and the other way round, because the Stokes theorem has no
logical direction. So all force fields $\vec{F}(\vec{r})$ with
$\nabla\times\vec{F}=0$ are conservative.

One can show easily (by explicitly carrying out the differentiation) that
force fields that can be written as the gradient of any potential $U(\vec{r})$
are curl-free and thus conservative:
\begin{equation}
 \nabla\times\left[\nabla U(\vec{r})\right]\equiv0
\end{equation}
This justifies the ansatz in (\ref{eq:gradientforce}) to write conservative
field as the gradient of a potential. 
Further, one can show that {\em every} sufficiently smooth field $\vec{F}$ with
$\nabla\times\vec{F}=0$ can be represented by a gradient of the form
(\ref{eq:gradientforce}).

\subsection{Total energy}\label{ssec:energyconservation}
For conservative fields, we can write the work $W_{12}$ for moving from
point 1 to 2 as a difference between the kinetic energy $T$ at the two points
due to (\ref{eq:workandT}), and as a difference of the potential energy at the
two endpoints according to (\ref{eq:potentialdifference}):
\begin{equation}\label{eq:differenceTandU}
W_{12}= T(\vec{r}_2)-T(\vec{r}_1)=U(\vec{r}_1)-U(\vec{r}_2)
\end{equation}
So the change in kinetic energy corresponds to the negative change in the
potential energy. Thus, it is reasonable to define a {\em total energy} $E$:
\begin{equation}
E:=T+U
\end{equation}
From (\ref{eq:differenceTandU}) it follows immediately that for a transition
from point 1 to point 2, the total energy does not change in a conservative
force field:
\begin{equation}
E(\vec{r}_1)=T(\vec{r}_1)+U(\vec{r}_1)=T(\vec{r}_2)+U(\vec{r}_2)=E(\vec{r}_2)
\end{equation}
Using this equality, it is easy to see that the total energy is the same for
every point along a path $\Gamma$, and that the temporal
derivative $dE/dt$ therefore vanishes --- or that the total energy $E$ {\em is
  conserved}. 

This derivation assumed that the force is not explicitly time dependent.  The
statement of conservation of the total energy can be slightly extended to
time-dependent potentials. To see that, we first use
(\ref{eq:kineticenergydifferential}) and transit from differentials to
temporal derivatives:
\begin{equation}
dT=\vec{F}\cdot d\vec{s}\,\Rightarrow
\frac{dT}{dt}=\vec{F}\cdot\frac{d\vec{r}}{dt}=\vec{F}\cdot\dot{\vec{r}}=\vec{F}\cdot\vec{v}
\end{equation}
The total temporal derivative of the potential experienced by a particle moving
along a path is composed by both the spatial dependency of $U$, and its
explicit time dependency:
\begin{eqnarray}
\frac{dU}{dt}&=&\sum\limits_i\frac{\partial U}{\partial
  x_i}\frac{dx_i}{dt}+\frac{\partial U}{\partial t}\nonumber\\
&=&(\nabla U)\cdot\vec{v}+\frac{\partial U}{\partial t}=
-\vec{F}\cdot\vec{v}+\frac{\partial U}{\partial t}
\end{eqnarray}
Adding the last two equations then yields the change of the total energy over
time:
\begin{equation}
\frac{dE}{dt}=\frac{dT}{dt}+\frac{dU}{dt}=\frac{\partial U}{\partial t}
\end{equation}
Again, if the potential $U$ does not explicitly depend on time, $dE/dt=0$, and
the {\em total energy is conserved\/}.

\subsubsection{Classification of problems}\label{sssec:classification}
The knowledge of the potential and the total energy of a system allows making
a lot of qualitative statements about the motion of a particle, even without
solving the full equation of motion. For this, we consider the potential $U(x)$
in a one-dimensional problem:

\begin{figure*}[h]
\centerline{\includegraphics{potential1.eps}}
\end{figure*}
First, the system can not have a total energy smaller than the lowest
potential $U(x)$ anywhere, because the kinetic energy $T=mv^2/2$ has to be
positive.  Thus, a system with total energy $E_0$ has no solution.

For $E=E_1$, there is a region between $x_1$ and $x_2$ where $T\ge0$. The
two positions $x_{1,2}$ are called {\em classical turning points\/}, because the
kinetic energy vanishes there, while the force does not. In such a scenario,
the physical system will oscillate between the classical turning points. The
solution of the equation of motion is referred to as a {\em bound solution\/},
because the motion is restricted to the finite interval $[x_1, x_2]$.

For a total energy $E=E_2$, there are two such intervals, $[x_3, x_4]$ and
$[x_5, x_6]$. The solution of the equation of motion will have two branches,
but there will be no transition between them. Any position $x$ of the system
outside these two intervals is not allowed, and said to be {\em forbidden by
  energy conservation\/}.

For $E=E_3$ (and assuming $U<E_3$ for $x>x_6$), the system has still
a classical turning point $x_7$, but is not bounded anymore. For
$t\rightarrow\pm\infty$, the system will evolve to $x\rightarrow+\infty$.

For the last case (assuming $U<E_4$ everywhere), there is no restriction
to the position $x$ of the system: a particle will come from
$x\rightarrow-\infty$ and evolve to $x\rightarrow+\infty$, or the other way
around. The last two cases are often found in {\em scattering  problems\/},
since the asymptotic position for $t\rightarrow\pm\infty$ is not finite,
whereas the interesting part of $U(\vec{x})$ is often limited to a small
region.

\subsubsection{Transit time}
The expression of a total energy in a conservative system can be used to
evaluate the time it takes a system to evolve between two points, without
having to solve the equation of motion explicitly. For a one-dimensional
problem and a given potential $U(x)$ and total energy $E$, one can invert the
expression for the total energy,
\begin{equation}
E=\frac{1}{2}mv^2+U(x)
\end{equation}
and obtain an expression for the velocity $v$ at a given position:
\begin{equation}
v=\pm\sqrt{\frac{2}{m}(E-U(x))}=\frac{dx}{dt}
\end{equation}
A single integration leads to an expression for the time difference between
two positions $x, x_0$ :
\begin{equation}
t-t_0=\int\limits_{t_0}^tdt'=\int\limits_{x_0}^x\frac{1}{v}\,dx'
=\int\limits_{x_0}^x\frac{dx'}{\sqrt{\frac{2}{m}(E-U(x))}}
\end{equation}
as long as the sign of the velocity does not change over the region
$[x_0,x]$. Such an expression can be used to evaluate the oscillation period
in an arbitrary potential leading to bound solutions.

\subsubsection{Potential of a harmonic oscillator}
A particularly important potential $U(x)$ in physics is that of a harmonic
oscillator which is characterized by a restoring force $\vec{F}(\vec{x})$
proportional to its displacement from a reference position
$\vec{x}=\vec{x}_0$. It can easily be seen that the potential
\begin{equation}\label{eq:harmoscenergy}
U(\vec{x})=\frac{1}{2}k(\vec{x}-\vec{x}_0)^2
\end{equation}
leads to the restoring force $\vec{F}=-k\,(\vec{x}-\vec{x}_0)$ seen in
\ref{sssec:harmonicosc1} for the one-dimensional case.
\begin{figure*}[h]
\centerline{\includegraphics{harmonicoscillator3.eps}}
\end{figure*}

As many local minima (with a non-vanishing second derivative $d^2U/dx^2$) can
be approximated by a parabola, many potentials with a local minimum lead to an
approximate harmonic oscillation around this minimum position $\vec{x}_0$.

\section{System of many particles}
So far, the mathematical formalism to describe the dynamics of a system was
restricted to a single particle. Many times, however, one is
interested in the dynamics of a system of many interacting
particles. These $N$ particles shall have individual masses $m_\alpha$ and
individual instantaneous positions $\vec{x}_\alpha$, where $\alpha$ is a
particle index.
\begin{figure*}[h]
\centerline{\includegraphics{manyparticles.eps}}
\end{figure*}

Often, one is not interested in the properties like position or velocity of
the individual particles, but in joint properties of the whole ensemble. It is
therefore  useful to define a few such quantities. Obviously, the {\em total
  mass} $M$ of the 
ensemble is given by
\begin{equation}\label{eq:totalmass}
  M:=\sum\limits_{\alpha=1}^Nm_\alpha
\end{equation}
With this, one can define a position vector $\vec{R}$ pointing to the {\em
  center of mass\/}
\begin{equation}\label{eq:centerofmass}
  \vec{R}:=\frac{1}{M}\sum\limits_{\alpha=1}^N m_\alpha \vec{x}_\alpha
\end{equation}
of the ensemble, which is an average position of all particles, weighted by
their mass. One can re-express the individual particle positions
$\vec{x}'_\alpha$ relative to
the center of mass:
\begin{equation}\label{eq:relativeposition}
  \vec{x}_\alpha=\vec{R}+\vec{x}'_\alpha
\end{equation}

\subsection{Center-of-mass motion}\label{ssec:COMmotion}
To obtain the Newtonian equations of motion for the ensemble, one needs to
know the force $\vec{F}_\alpha$ acting each particle $\alpha$. It is
convenient to split this force into a component $\vec{f}_\alpha$ caused by the
interaction between the particles in the ensemble, and a component
$\vec{F}_\alpha^{ext}$ due to external interactions:
\begin{equation}\label{eq:splitforce1}
  \vec{F}_\alpha=\vec{F}_\alpha^{ext}+\vec{f}_\alpha\quad{\rm with}\quad
\vec{f}_\alpha=\sum\limits_{\beta\atop\beta\neq\alpha}\vec{f}_{\alpha\beta}
\end{equation}
The latter sum considers interactions between particle $\alpha$ with all other
particles $\beta$. For central forces, like gravitational attraction or
Coulomb interaction, these internal forces are symmetric according to Newton's
third law:
\begin{equation}\label{eq:internalforcesymmetry}
\vec{f}_{\alpha\beta}=-\vec{f}_{\beta\alpha}
\end{equation}

The equation of motion for the whole ensemble is then simply the set of
equations of motion for the individual particles for all $\alpha$:
\begin{equation}\label{eq:eqmotionensemble}
m\ddot\vec{x}_\alpha=\vec{F}_\alpha^{ext}+\sum\limits_{\beta\atop\beta\neq\alpha}\vec{f}_{\alpha\beta}\quad\alpha=1\ldots N
\end{equation}

The summation over all equations in (\ref{eq:eqmotionensemble}) leads on the
left side to
\begin{equation}
\sum\limits_\alpha m\ddot\vec{x}_\alpha=\frac{d^2}{dt^2}\sum\limits_\alpha m\vec{x}_\alpha=M\ddot\vec{R}\,.
\end{equation}
On the right side, we can define a {\em total external force}
\begin{equation}
\vec{F}^{ext}:=\sum\limits_\alpha\vec{F}^{ext}_\alpha
\end{equation}
The sum over all internal forces $\vec{f}_{\alpha\beta}$ vanishes,
\begin{equation}
\sum\limits_{\alpha,\beta\atop\alpha\neq\beta}\vec{f}_{\alpha\beta}=0\,,
\end{equation}
because in the double summation over $\alpha$ and $\beta$, each term
$\vec{f}_{\alpha\beta}$ gets canceled out by the term
$\vec{f}_{\beta\alpha}$. 
Thus, the sum over all equations of motion (\ref{eq:eqmotionensemble}) leads
to an equation of motion for the center of mass $\vec{R}$ of the ensemble
where all internal forces between the particles vanish:
\begin{equation}
M\ddot\vec{R}=\vec{F}^{ext}
\end{equation}
This equation of motion has the same form as the one for a single particle.
Similarly to the definition of a momentum for an individual particle, one
can define a {\em total linear momentum} 
\begin{equation}
\vec{P}:=\sum\limits_\alpha m_\alpha\dot\vec{x}_\alpha=M\dot\vec{R}\,,
\end{equation}
with a time derivative that is only determined by the external force:
\begin{equation}
\dot\vec{P}=M\ddot\vec{R}
%= \frac{d}{dt}\sum\limits_\alpha m_\alpha\dot\vec{x}_\alpha
%=\sum\limits_\alpha m\ddot\vec{x}_\alpha
=\vec{F}^{ext}
\end{equation}
This allows considering the motion of a whole ensemble (or more precisely,
collective properties of the ensemble like $\vec{R}$ and $\vec{P}$) in the
same way as a single point with mass $M$.

\subsection{Total angular momentum}\label{ssec:angularmomentum}
The individual particles in the ensemble all have a well-defined angular
momentum with respect to a coordinate origin $O$ as defined in
(\ref{eq:angularmomentumdef}),
\begin{equation}
\vec{l}_\alpha=\vec{x}_\alpha\times\vec{p}_\alpha=\vec{x}_\alpha\times m_\alpha\dot\vec{x}_\alpha
\end{equation}
The {\em total angular momentum} of the ensemble is then given by
\begin{equation}
\vec{L}:=\sum\limits_\alpha\vec{l}_\alpha
% =\sum\limits_\alpha\vec{x}_\alpha\times m_\alpha\dot\vec{x}_\alpha
\end{equation}
This definition can be expressed in terms of the center-of-mass position
$\vec{R}$ of the ensemble, and the individual displacements $\vec{x}'_\alpha$
(\ref{eq:relativeposition}):
\begin{eqnarray}\label{eq:totalangular2}
\vec{L}&=&\sum\limits_\alpha(\vec{R}+\vec{x}'_\alpha)\times
m_\alpha(\dot{\vec{R}}+\dot{\vec{x}}'_\alpha)\nonumber\\
&=&\sum\limits_\alpha m_\alpha\left[\vec{R}\times\dot\vec{R} +
\vec{x}'_\alpha\times\dot\vec{x}'_\alpha\right]\nonumber\\
& &\quad+ 
\vec{R}\times\left(\sum\limits_\alpha m_\alpha\dot\vec{x}'_\alpha\right)
+\left(\sum\limits_\alpha m_\alpha\vec{x}'_\alpha\right)\times\dot\vec{R}
\end{eqnarray}
The expressions in the parentheses are center of mass position of an ensemble
with respect to the center of mass position, and
\begin{equation}
\sum\limits_\alpha m_\alpha\vec{x}'_\alpha=\sum\limits_\alpha m_\alpha(\vec{x}_\alpha-\vec{R})=\sum\limits_\alpha m_\alpha\vec{x}_\alpha-\sum\limits_\alpha m_\alpha\vec{R}=M\vec{R}-M\vec{R}=0\,,
\end{equation}
so the last two terms in (\ref{eq:totalangular2}) vanish. With
$\vec{P}=M\dot\vec{R}$, this leads to
\begin{equation}\label{eq:totalangular3}
\vec{L}=\vec{R}\times\vec{P}+\sum\limits_\alpha\vec{x}'_\alpha\times\vec{p}'_\alpha\quad(\,{\rm with}\,\vec{p}'_\alpha:=m_\alpha\dot\vec{x}'_\alpha\,)\,.
\end{equation}
The total angular momentum is therefore the sum of a contribution from the
center of mass, and a contribution of the total angular momentum with respect
to the center of mass of the ensemble. This will be become important in the
Huygens-Steiner theorem for moments of inertia.

To understand the dynamics of the total angular momentum in a differential
equation for $\vec{L}$ similar to (\ref{eq:angularmomemtumchange}), we first
consider the temporal derivative of the individual angular momenta:
\begin{equation}
\dot\vec{l}_\alpha=\frac{d}{dt}(\vec{x}_\alpha\times\vec{p}_\alpha)=\underbrace{\dot\vec{x}_\alpha\times\vec{p}_\alpha}_{=0}+\vec{x}_\alpha\times\dot\vec{p}_\alpha
\end{equation}
The first term vanishes because $\vec{p}_\alpha=m_\alpha\dot\vec{x}_\alpha$,
and cross products of parallel vectors are zero. By using
$\dot\vec{p}_\alpha=\vec{F}_\alpha$ and the decomposition
(\ref{eq:splitforce1}) one obtains
\begin{equation}
\dot\vec{l}_\alpha=\vec{x}_\alpha\times\left(\vec{F}_\alpha^{ext}+\sum\limits_{\beta\atop\alpha\neq\beta}\vec{f}_{\alpha\beta}\right)
\end{equation}
and for the derivative of the total angular momentum 
\begin{equation}
\dot\vec{L}=\sum\limits_\alpha\dot\vec{l}_\alpha=
\sum\limits_\alpha\left(\vec{x}_\alpha\times\vec{F}_\alpha^{ext}\right)+
\sum\limits_{\alpha,\beta\atop\alpha\neq\beta}\vec{x}_\alpha\times\vec{f}_{\alpha\beta}\,.
\end{equation}
To see that the last term vanishes, we reorder the summation
\begin{equation}
\sum\limits_{\alpha,\beta\atop\alpha\neq\beta}A_{\alpha\beta}=\sum\limits_{\alpha,\beta\atop\alpha<\beta}A_{\alpha\beta}+\sum\limits_{\alpha,\beta\atop\alpha>\beta}A_{\alpha\beta}=\sum\limits_{\alpha,\beta\atop\alpha<\beta}\left(A_{\alpha\beta}+A_{\beta\alpha}\right)\,.
\end{equation}
With $A_{\alpha\beta}=\vec{x}_\alpha\times\vec{f}_{\alpha\beta}$ and using the
symmetry $\vec{f}_{\alpha\beta}=-\vec{f}_{\beta\alpha}$ of the inter-particle forces (\ref{eq:internalforcesymmetry}), one
finds
\begin{equation}\label{eq:sumrelativetorque}
\sum\limits_{\alpha,\beta\atop\alpha\neq\beta}\vec{x}_\alpha\times\vec{f}_{\alpha\beta}=
\sum\limits_{\alpha,\beta\atop\alpha<\beta}(\vec{x}_\alpha-\vec{x}_\beta)\times\vec{f}_{\alpha\beta}\,.
\end{equation}
For central inter-particle forces, $\vec{f}_{\alpha\beta}$ is parallel to the
distance vector $\vec{x}_\alpha-\vec{x}_\beta$ between them, so the cross
product vanishes; therefore, the whole sum (\ref{eq:sumrelativetorque})
vanishes. With the torque
$\vec{N}_\alpha^{ext}:=\vec{x}_\alpha\times\vec{F}_\alpha^{ext}$ on particle $\alpha$ according to definition (\ref{eq:torquedefinition}) caused by the external force
$\vec{F}_\alpha^{ext}$, the change of the total angular momentum with time is
given by
\begin{equation}\label{eq:changeofangularmomentum}
\dot\vec{L}=\sum\limits_\alpha\vec{N}_\alpha^{ext}=:\vec{N}^{ext}\,.
\end{equation}
Analog to the single particle case in (\ref{eq:angularmomemtumchange}), the
total angular momentum of the ensemble with respect to an origin $O$ remains
constant if the {\em total external torque} $\vec{N}^{ext}$ with respect to $O$
vanishes.

\subsection{Work and total kinetic energy}\label{ssec:totalkineticenergy}
Similar to the single particle case, we consider the sum of the work for a
transition of the ensemble between two configurations $a$ and $b$:
\begin{equation}\label{eq:ensemblework}
W_{ab}=\sum\limits_\alpha\int\limits_a^b\vec{F}_\alpha\cdot d\vec{x}_\alpha
\end{equation}
Here, the configurations $a$ and $b$ represent sets of positions
$\{\vec{x}_1,\vec{x}_2,\ldots\}$ for all particles evolving in time. The
integral is a path integral for the trajectory of particle $\alpha$ in this
transition. In exactly the same way as for the single particle case in
section~\ref{ssec:kineticenergy}, one replaces the kernel in the integral via
\begin{equation}
\vec{F}_\alpha\cdot d\vec{x}_\alpha=dT_\alpha\quad{\rm
  with}\quad T_\alpha:=\frac{1}{2}m_\alpha\vec{v}_\alpha^2\,,
\end{equation}
and can express the total work as
\begin{equation}\label{eq:totalkineticwork}
W_{ab}=\sum\limits_\alpha\int\limits_a^bdT_\alpha=T_b-T_a\,,
\end{equation}
with a {\em total kinetic energy\/} $T$ of a configuration $a$ or $b$ given by
the sum of the individual kinetic energies,
\begin{equation}\label{eq:totalkineticenergy}
T:=\sum\limits_\alpha
T_\alpha=\sum\limits_\alpha\frac{1}{2}m_\alpha\vec{v}_\alpha^2\,.
\end{equation}
By splitting the velocity of the individual particles into a center-of-mass
velocity $\dot\vec{R}$, and a relative velocity $\vec{x}'_\alpha$  with
respect to the center-of-mass,
\begin{equation}
\vec{v}_\alpha^2=\dot\vec{x}^2_\alpha=(\dot\vec{x}'_\alpha+\dot\vec{R})\cdot(\dot\vec{x}'_\alpha+\dot\vec{R})=\vec{v}'^2_\alpha+2\dot\vec{x}_\alpha\cdot\dot\vec{R}+\dot\vec{R}\cdot\dot\vec{R}\,,
\end{equation}
the total kinetic energy becomes
\begin{eqnarray}
T&=&\sum\limits_\alpha\frac{1}{2}m_\alpha
v'^2_\alpha+\sum\limits_\alpha\frac{1}{2}m_\alpha\dot\vec{R}^2+\dot\vec{R}\frac{d}{dt}\underbrace{\sum\limits_\alpha
  m_\alpha\vec{x}'_\alpha}_{=0}\nonumber\\
&=&\sum\limits_\alpha\frac{1}{2}m_\alpha v'^2_\alpha+\frac{1}{2}M\dot\vec{R}^2\,.
\end{eqnarray}
So the total kinetic energy is a sum of the kinetic energy of the relative
motion of the particles with respect to each other, and the kinetic energy of the center-of-mass motion in
the form of a single particle with the total mass $M$ of the ensemble.

\subsection{Total potential energy}
Analog to the evaluation of the work integral in section
\ref{ssec:conservativeforce} for a single particle in a conservative force
field through a potential, several potentials can be used to describe
conservative forces in a particle ensemble. First, the forces
on particle $\alpha$ in (\ref{eq:ensemblework}) get split up in external and
internal contributions according to (\ref{eq:splitforce1}):
\begin{equation}\label{eq:manywork2}
W_{ab}=\sum\limits_\alpha\int\limits_a^b\vec{F}_\alpha^{ext}\cdot d\vec{x}_\alpha+
\sum\limits_{\alpha,\beta\atop\alpha\neq\beta}\int\limits_a^b\vec{f}_{\alpha\beta}\cdot
d\vec{x}_\alpha
\end{equation}
Conservative external and internal forces can be written as
\begin{equation}
\vec{F}_\alpha^{ext}=-\nabla_\alpha U_\alpha\,,\quad
\vec{f}_{\alpha\beta}=-\nabla_\alpha \overline{U}_{\alpha\beta}\,,
\end{equation}
where $\nabla_\alpha$ denotes a differentiation with respect to the
coordinates $\vec{x}_\alpha$ of particle $\alpha$. The two potentials
are different functions, where $U_\alpha$ represents external potentials,
while the $\overline{U}_{\alpha\beta}$ 
describes the interaction between particles $\alpha$ and $\beta$ within the
ensemble.

The first term in (\ref{eq:manywork2}) is completely analog to the single
particle case:
\begin{equation}
\sum\limits_\alpha\int\limits_a^b\vec{F}_\alpha^{ext}\cdot d\vec{x}_\alpha=
-\sum\limits_\alpha\int\limits_a^b(\nabla_\alpha U_\alpha)\cdot d\vec{x}_\alpha=
-\left.\left(\sum\limits_\alpha U_\alpha\right)\right|_a^b
\end{equation}
For the second term in (\ref{eq:manywork2}), the sum gets split up over half
of the combinations $\alpha,\beta$ like in (\ref{eq:sumrelativetorque}),
\begin{equation}
\sum\limits_{\alpha,\beta\atop\alpha\neq\beta}\vec{f}_{\alpha\beta}\cdot d\vec{x}_\alpha=
\sum\limits_{\alpha,\beta\atop\alpha<\beta}(\vec{f}_{\alpha\beta}\cdot
d\vec{x}_\alpha+\vec{f}_{\beta\alpha}\cdot
d\vec{x}_\beta)=\sum\limits_{\alpha,\beta\atop\alpha<\beta}\vec{f}_{\alpha\beta}\cdot d(\vec{x}_\alpha-\vec{x}_\beta)\,.
\end{equation}

\noindent
Formally, the total differential $d\overline{U}_{\alpha\beta}$ can be written
as
\begin{eqnarray}\label{eq:totaldifferentialu}
d\overline{U}_{\alpha\beta}&=&\sum\limits_i
\frac{\partial\overline{U}_{\alpha\beta}}{\partial x_{\alpha,i}}\,dx_{\alpha,i}+
\frac{\partial\overline{U}_{\alpha\beta}}{\partial x_{\beta,i}}\,dx_{\beta,i}\nonumber\\
&=&(\nabla_\alpha\overline{U}_{\alpha\beta})\cdot d\vec{x}_\alpha+
(\nabla_\beta\overline{U}_{\alpha\beta})\cdot d\vec{x}_\beta\,.
\end{eqnarray}
For inter-particle central forces, the potential $\overline{U}_{\alpha\beta}$
only depends on the modulus of the distance, $|\vec{x}_\alpha-\vec{x}_\beta|$,
therefore, $\overline{U}_{\alpha\beta}=\overline{U}_{\beta\alpha}$. Then, 
\begin{equation}
\nabla_\beta\overline{U}_{\alpha\beta}=\nabla_\beta\overline{U}_{\beta\alpha}=-\vec{f}_{\beta\alpha}=\vec{f}_{\alpha\beta}\,,
\end{equation}
and the total differential $d\overline{U}_{\alpha\beta}$ in
(\ref{eq:totaldifferentialu}) becomes
\begin{equation}
d\overline{U}_{\alpha\beta}=-\vec{f}_{\alpha\beta}\cdot d\vec{x}_\alpha+
\vec{f}_{\alpha\beta}\cdot d\vec{x}_\beta=
-\vec{f}_{\alpha\beta}\cdot d(\vec{x}_\alpha-\vec{x}_\beta)
\end{equation}
With this, the second term in the total work in (\ref{eq:manywork2}) can
be written as 
\begin{equation}
\sum\limits_{\alpha,\beta\atop\alpha\neq\beta}\int\limits_a^b\vec{f}_{\alpha\beta}\cdot
d\vec{x}_\alpha=
-\sum\limits_{\alpha,\beta\atop\alpha<\beta}\int\limits_a^bd\overline{U}_{\alpha\beta}=
-\left.\sum\limits_{\alpha,\beta\atop\alpha<\beta}\overline{U}_{\alpha\beta}\right|_a^b\,,
\end{equation}
so the complete expression for the total work (\ref{eq:manywork2}) is
\begin{equation}\label{eq:totalpotentialwork}
W_{ab}=-\left.\left(\sum\limits_\alpha U_\alpha\right)\right|_a^b
-\left.\sum\limits_{\alpha,\beta\atop\alpha<\beta}\overline{U}_{\alpha\beta}\right|_a^b
=-\left.U\right|_a^b=U_a-U_b\,,
\end{equation}
where the last $U$ in the expression is a {\em total potential energy}
\begin{equation}\label{eq:totalpotentialenergy}
U:=\underbrace{\sum\limits_\alpha U_\alpha}_{\rm
  external}+\underbrace{\sum\limits_{\alpha,\beta\atop\alpha<\beta}\overline{U}_{\alpha\beta}}_{\rm internal}
\end{equation}
for a given system state defined by positions $\{\vec{x}_1,\vec{x}_2,\ldots\}$
of all particles. 

\subsection{Energy conservation}\label{ssec:energyconservationensemble}
As before in case of a single particle, the work in a configuration change can
be written both as a difference in total kinetic and total potential
energy. Combining (\ref{eq:totalkineticwork}) and
(\ref{eq:totalpotentialwork}) gives
\begin{equation}
W_{ab}=T_b-T_a=U_a-U_b\,,
\end{equation}
or after reordering
\begin{equation}
T_a+U_a=T_b+U_b\,,
\end{equation}
which means the total energy $E:=T+U$ of the many-particle system is conserved
in an evolution between states $a$ and $b$.

It should be noted that for {\em rigid bodies}, the second
term in (\ref{eq:manywork2}) will vanish because the inter-particle distances
$||\vec{x}_\alpha-\vec{x}_\beta||$ will not change in a system state change
$a\rightarrow b$. Thus, the
contribution (\ref{eq:totalpotentialenergy}) due to the internal interactions
stay constant, and need not to be evaluated explicitly. Then, the total
kinetic and total potential energy in the system are just the sum of the
individual contributions from the participating particles.

\section{Lagrangian mechanics - first approach}
So far, the equations of motion have been derived from Newton's laws, and
required the knowledge of forces for all particles. This can sometimes become
challenging, in particular in cases where forces need to be
considered that are only there to meet the constraints.  In general, the
equations of motion of a system following Newtonian mechanics can be written as
\begin{equation}\label{eq:newtonequations}
m_k\,\ddot{x_k}=F_k(\{x_l, \dot{x_l}\}, t)\,.
\end{equation}
This is a system of differential equations for all $x_k$, where $k$ indexes
both the coordinate components (like $x,y,z$) and a particle index in a
many particle system: a system of two particles moving in 3-dimensional space
would lead to 6 differential equations. The notation $\{x_l,
\dot{x_l}\}$ indicates that in principle, all forces $F_k$ can
depend on coordinates $x_l$ and velocities $\dot{x_l}$ of all
particles. Additionally they can explicitly depend on time $t$. 

Similarly, the total kinetic energy of the system
(\ref{eq:totalkineticenergy}) can be written as
\begin{equation}\label{eq:lagKineticenergy}
T=T(\{\dot{x}_l\})=\sum\limits_kT_k=\sum\limits_k\frac{1}{2}m_k\dot{x}_k^2\,.
\end{equation}
The Cartesian momentum component for coordinate index $k$ is then given by 
\begin{equation}\label{eq:momentumfromT}
p_k=m_k\dot x_k=\frac{\partial T}{\partial \dot{x_k}}\,,
\end{equation}
because differentiation of the sum in (\ref{eq:lagKineticenergy}) with respect
to $\dot{x}_l$ vanish for $l\neq k$. The left side of (\ref{eq:newtonequations}) can be expressed
by the total temporal derivative of (\ref{eq:momentumfromT}),
\begin{equation}\label{eq:newtonequations2}
m_k\,\ddot x_k=\dot{p}_k=\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{x_k}}\right)=
F_k(\{x_l, \dot{x_l}\}, t)\,.
\end{equation}
For conservative forces, one can write $F_k=-\partial U/\partial x_k$, with a
total potential energy $U$ for the whole system. This is compatible with the
definition in (\ref{eq:totalpotentialenergy}). Then, the equations of motion become
\begin{equation}
\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{x_k}}\right)=-\frac{\partial
U}{\partial x_k}\quad{\rm for~all~}k.
\end{equation}
With the definition of a so-called {\em Lagrange function} $L:=T-U$, this can
be written as
\begin{equation}\label{eq:LagrangeEoM}
\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x_k}}\right)-\frac{\partial
L}{\partial x_k}=0\quad{\rm for~all~}k\,,
\end{equation}
because for the differentiation with respect to $\dot{x_k}$, there is no
contribution from $U$, and similarly, for the differentiation  with respect to
$x_k$, there is no contribution from $T$. The set (\ref{eq:LagrangeEoM}) is
called {\em Lagrange equations of motion} of a physical system, and are
equivalent to the equation of motion (\ref{eq:newtonequations}) in Newtonian
mechanics.

So far, there is no obvious advantage of this method for obtaining the
equations of motion. However, it will simplify the treatment of systems,
because these equations of motion take the same form in also in general,
possibly non-Cartesian coordinates that reflect better the symmetry of a
system.

\subsection{Example: harmonic oscillator}\label{ssec:harmonicosc}
We consider again the one-dimensional harmonic oscillator. The
kinetic energy is given by $T=mv^2/2$, and the potential energy according to
(\ref{eq:harmoscenergy}) by $U=kx^2$ for an equilibrium position
$x_0=0$. Then, the Lagrange function becomes
\begin{equation}
L=T-U=\frac{1}{2}m\,\dot{x}^2-\frac{1}{2}k\,x^2
\end{equation}
The corresponding Lagrange equation is then
\begin{equation}
\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x_k}}\right)-\frac{\partial
L}{\partial x_k}=
\frac{d}{dt}\left(m\,\dot{x}\right)+kx=m\,\ddot{x}+kx=0\,,
\end{equation}
which is equivalent to (\ref{eq:harmosc2}).


\subsection{Generalization to some non-conservative forces}\label{ssec:nonconservative}
With the Lagrange formalism, it is also possible to consider some
non-conservative 
velocity-dependent forces, as long as they can be written as
\begin{equation}\label{eq:moregeneralpotential}
F_k=F_k(\{x_l,\dot{x}_l\},t)=-\frac{\partial V}{\partial
  x_k}+\frac{d}{dt}\frac{\partial V}{\partial\dot{x}_k}\,
\end{equation}
where $V=V(\{x_l,\dot{x}_l\},t)$ is a velocity-dependent pseudo-potential.
With a Lagrange function $L=T-V$, the force on the left side
of the Newtonian equation of motion (\ref{eq:newtonequations2}) can be written
as 
\begin{equation}
m_k\ddot{x}_k = \frac{d}{dt}\frac{\partial T}{\partial \dot{x}_k}=\frac{d}{dt}\frac{\partial
  L}{\partial \dot{x}_k}+\frac{d}{dt}\frac{\partial
  V}{\partial \dot{x}_k}
\end{equation}
With the right side of the Newtonian equation of motion
(\ref{eq:newtonequations2}) taken from (\ref{eq:moregeneralpotential}), and
observing that $\partial L/\partial x_k=-\partial V/\partial x_k$ because $T$
does not depend on $x_k$, one finds
\begin{equation}
\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_k}+
\frac{d}{dt}\frac{\partial V}{\partial \dot{x}_k}=
 -\frac{\partial V}{\partial x_k} + \frac{d}{dt}\frac{\partial
   V}{\partial\dot{x}_k}=
\frac{\partial L}{\partial x_k} + \frac{d}{dt}\frac{\partial
   V}{\partial\dot{x}_k}\,.
\end{equation}
By removing the second derivatives of $V$ on both sides, one ends up with the
Lagrange equation of motion of the form (\ref{eq:LagrangeEoM}).

This expands the usefulness of the Lagrange formalism significantly beyond
conservative forces. An important example of such a force is the one felt by a
charged particle in the presence of both an electrical and magnetic field.

\subsubsection{Lagrange function for a charge under electromagnetic forces}
A particle with a point charge $q$ is subject to forces both due to electrical
and magnetic fields:
\begin{equation}\label{eq:elmforce}
\vec{F}=q\left(\vec{E}+\vec{v}\times\vec{B}\right)
\end{equation}
The first part is due to the electric field vector $\vec{E}$, the second part
is the Lorentz force due to the presence of a magnetic field $\vec{B}$. 

In order to show that such an interaction can be written in a Lagrangian form,
we make use from a result from electrodynamics that the electric and
magnetic fields can be derived from two potentials,
\begin{equation}\label{eq:fieldfrompotentials}
\vec{E}=-\nabla \Phi-\frac{\partial}{\partial t}\vec{A}\,,\quad{\rm and}\quad\vec{B}=\nabla\times\vec{A}\,,
\end{equation}
where $\Phi=\Phi(\vec{x},t)$ is the scalar electrical potential, and
$\vec{A}=\vec{A}(\vec{x},t)$ the so-called vector potential. We will show that
a ``pseudopotential'' energy 
\begin{equation}\label{eq:pseudopotential}
V=V(\vec{x},\dot\vec{x},t)=q\left[\Phi(\vec{x},t)-\dot\vec{x}\cdot\vec{A}(\vec{x},t)\right]
\end{equation}
can reproduce the electromagnetic force (\ref{eq:elmforce}) via
(\ref{eq:moregeneralpotential}). For this, we first rewrite (\ref{eq:elmforce})
in terms of individual components $k$, and express the fields $\vec{E},
\vec{B}$ by the potentials $\Phi, \vec{A}$ according to (\ref{eq:fieldfrompotentials}):
\begin{eqnarray}
F_k&=&q\left[E_k + \sum\limits_{l,m} \epsilon_{klm}\dot{x}_lB_m\right]\nonumber\\
&=&q \left[\left(-\frac{\partial\Phi}{\partial x_k}-\frac{\partial A_k}{\partial
    t}\right) + \sum\limits_{l,m}\epsilon_{klm}
\dot{x}_l\left(\sum\limits_{p,q}\epsilon_{mpq}\frac{\partial A_q}{\partial
    x_p}\right)\right]\nonumber\\
&=&q \left[\left(-\frac{\partial\Phi}{\partial x_k}-\frac{\partial A_k}{\partial
    t}\right) +
\sum\limits_{l,p,q}\left(\sum\limits_m\epsilon_{klm}\epsilon_{mpq}\right)\dot{x}_l\frac{\partial
  A_q}{\partial x_p}\right]
\end{eqnarray}
By using an identity for the sum over the Levi-Civita symbols,
\begin{equation}
\sum\limits_m\epsilon_{klm}\epsilon_{mpq}=\sum\limits_m\epsilon_{mkl}\epsilon_{mpq}=\delta_{kp}\delta_{lq}-\delta_{kq}\delta_{lp}\,,
\end{equation}
the summations over $p$ and $q$ can be removed:
\begin{equation}\label{eq:forcefromelmfields}
F_k=q \left[\left(-\frac{\partial\Phi}{\partial x_k}-\frac{\partial A_k}{\partial
    t}\right)+
\sum\limits_l\dot{x}_l\left(\frac{\partial A_l}{\partial x_k}-
  \frac{\partial A_k}{\partial x_l}\right)\right]
\end{equation}
In a next step, we derive the force via (\ref{eq:moregeneralpotential}) from
the pseudopotential $V$ in (\ref{eq:pseudopotential}):
\begin{eqnarray}
F_k&=&-\frac{\partial V}{\partial
  x_k}+\frac{d}{dt}\frac{\partial V}{\partial\dot{x}_k}\nonumber\\
&=&q\left[-\frac{\partial\Phi}{\partial x_k}+\sum\limits_l\dot{x}_l\frac{\partial
    A_l}{\partial x_k} -\frac{d}{dt}A_k\right]\nonumber\\
&=&q\left[-\frac{\partial\Phi}{\partial x_k}+\sum\limits_l\dot{x}_l\frac{\partial
    A_l}{\partial x_k}-\frac{\partial A_k}{\partial t}-\sum\limits_l\frac{\partial A_k}{\partial x_l}\dot{x}_l\right]\,,
\end{eqnarray}
which is exactly the same as the force derived from the
fields in (\ref{eq:forcefromelmfields}). Therefore, the forces
(\ref{eq:elmforce}) can be expressed via the pseudopotential (\ref{eq:pseudopotential}).

As a consequence, the Lagrange function or {\em Lagrangian} for a charged
particle in time-dependent electromagnetic fields is given by 
\begin{equation}\label{eq:lagrangianchargedparticle}
L=T-V=\frac{1}{2}m\,\dot\vec{x}^2-q\left[\Phi(\vec{x},t)-\dot\vec{x}\cdot\vec{A}(\vec{x},t)\right]\,,
\end{equation}
and an equation of motion for it can be obtained via the Lagrange
equations.

\section{Lagrangian mechanics from Hamilton's principle}
In the previous section, the dynamics of a physical system was expressed not via
equations of motion provided by Newton's laws, but via the definition of a the
Lagrange function $L=L(\{x_l,\dot{x}_l\},t)=T-U$, and a set of differential
equations, the so-called Lagrange equations (\ref{eq:LagrangeEoM}). 

Since the Lagrange equations are just a mathematical prescription how to
obtain equations of motion for a system, all the information on how a system
evolves is encoded in the form of the Lagrange function. So far, we used
Newton's laws, which use forces as their central concept, to 
derive the Lagrange equations. However, the forces do not appear in the final
equation of motion.
Thus, several attempts were made to derive the equations of motion from
different principles, and that do not require the notion of forces.

A long-time popular concept is the principle of minimizing a
certain quantity: For example, to explain how light is reflected off a mirror, 
Hero\footnote{\href{https://en.wikipedia.org/wiki/Hero_of_Alexandria}{Hero
    of Alexandria}, $\approx$10-70, in year 60} postulated that the light takes
the shortest path between two points. About 1000 years later, this 
principle was extended by Ibn al-Haytham\footnote{\href{https://en.wikipedia.org/wiki/Alhazen}{Ibn
    al-Haytham} from Cairo, 965-1040} to describe the refraction of light as
well, postulating the principle that the path of the light is determined by
the shortest time to move between two points. This was stated in a modern from
by Fermat\footnote{\href{https://en.wikipedia.org/wiki/Pierre_de_Fermat}{Pierre
    de Fermat}, Toulouse, France,  1601-1665} and is known as {\em
  \href{https://en.wikipedia.org/wiki/Fermat's_principle}{Fermat's principle}}.

After a number of attempts to come up with such a principle determining the
mechanics by physicists over the time,
Hamilton\footnote{\href{https://en.wikipedia.org/wiki/William_Rowan_Hamilton}{Sir
    William Rowan Hamilton, 1805-1865}} announced in 1834/35 the {\em dynamic
  principle}, often referred to as {\em Hamilton's principle}, that does
exactly this. It basically states:\vspace*{1em}

\hfill\begin{minipage}{\dimexpr\textwidth-1cm}{\em Of all possible paths a system may
    move from one point to another within a specified time interval, the
    actual path followed is the one that minimizes the integral over the
    difference between kinetic and potential energies.}
\end{minipage}

A mathematical formulation of this is often written as 
\begin{equation}\label{eq:hamiltonprinciple}
\delta S =0\,,\quad{\rm with}\quad S:=\int\limits_{t_1}^{t_2}(T-U)\,dt=\int\limits_{t_1}^{t_2}\!L\,dt\,,
\end{equation}
where the the time integral $S$ is referred to as {\em action} of a physical
system when evolving between two different states at times $t_1$ and $t_2$.
The delta symbol here refers somewhat vaguely to a {\em variation} of a
quantity, which tries to capture what is meant by ``the trajectory chosen by
nature of the system will make $S$ extremal''. The quantity $\delta S$  is the
change of $S$ with a variation of the final trajectory, and should vanish for
the extremal path -- in a similar way that  the change $df$ of a function
$f(x)$ vanishes with variation  $x\rightarrow x+dx$ near a minimum or maximum
of $f(x)$. 
The mathematical discipline of {\em variational calculus} tries to solve
exactly this problem.

\subsection{Elements of variational calculus - Euler equation}\label{ssec:EulerEquations}
To solve the variational problem of a kind proposed by Hamilton's principle,
we first consider a function $y(x)$, and a recipe to obtain a number $J$ from
this function,
\begin{equation}\label{eq:costfunctional}
J=\int\limits_{x_1}^{x_2}f(y(x), y'(x), x)\,dx\,,
\end{equation}
where $f$ is a function that depends on three parameters: the function $y(x)$
itself, its derivative $y'(x)$ with respect to the function parameter $x$, and 
$x$ itself. Such a function is referred to as a {\em functional}. Examples for
such functions would be
\begin{equation}
f=y\sqrt{(1+y'^2)}\quad{\rm or}\quad f=\sqrt{\frac{1+y'^2}{x}}\,,
\end{equation}
where the first one does not explicitly depend on $x$, and the second one not
explicitly on $y$.

To find a condition on $y(x)$ that minimizes or maximizes the value $J$, we consider a small deviation of $y(x)$ from that optimum in
the form of a variation
\begin{equation}\label{eq:smalldeviation1}
y(\alpha,x)=y(x)+\alpha\eta(x)\,,
\end{equation}
where $\alpha$ is a parameter to gradually add  the ``deviation'' $\eta(x)$ to
the optimal solution $y(x)$. At the end points $x_{1,2}$, the deviation should
vanish, $\eta(x_1)=\eta(x_2)=0$.

\begin{figure*}[h]
\centerline{\includegraphics{variation1.eps}}
\end{figure*}

Now, the functional $J$ becomes a function of the perturbation parameter
$\alpha$,
\begin{equation}
J(\alpha)=\int\limits_{x_1}^{x_2}f(y(\alpha,x), y'(\alpha,x),x)\,dx\,,
\end{equation}
For a function $y(x)$ that makes $J$ extremal, one would require the condition
\begin{equation}\label{eq:variationcondition}
\left.\frac{\partial J}{\partial\alpha}\right|_{\alpha=0}\stackrel{!}{=}0\quad{\rm
for\,\,all}\quad \eta(x)\,,
\end{equation}
because a small change from the optimal $y(x)$ will not change the value of
$J$ at the optimum. Condition (\ref{eq:variationcondition}) will now lead to a
way to construct $y(x)$:
\begin{eqnarray}
\frac{\partial J}{\partial\alpha}&=&
\int\limits_{x_1}^{x_2}\left(\frac{\partial f}{\partial y}\frac{\partial y}{\partial\alpha}+\frac{\partial f}{\partial
  y'}\frac{\partial y'}{\partial\alpha}\right)\,dx
=\int\limits_{x_1}^{x_2}\Bigg(\frac{\partial f}{\partial y}
  \underbrace{\frac{\partial y}{\partial\alpha}}_{=\eta(x)}+
  \frac{\partial f}{\partial y'}
  \underbrace{\frac{\partial^2 y}{\partial\alpha\partial
      x}}_{=\frac{d\eta(x)}{dx}}\Bigg)\,dx\nonumber\\
&=&\int\limits_{x_1}^{x_2}\Bigg(\frac{\partial f}{\partial y}\eta(x)\Bigg)\,dx+
\int\limits_{x_1}^{x_2}\Bigg(\frac{\partial f}{\partial y'}\frac{d\eta(x)}{dx}\Bigg)\,dx
\end{eqnarray}
The last part can be integrated using $u=\partial f/\partial y',
v=\eta(x)$, and $\int uv'=uv-\int u'v$:
\begin{eqnarray}
\frac{\partial J}{\partial\alpha}&=&
\int\limits_{x_1}^{x_2}\Bigg(\frac{\partial f}{\partial y}\eta(x)\Bigg)\,dx+
\underbrace{\frac{\partial f}{\partial y'}\eta(x)\Bigg|_{x_1}^{x_2}}_{=0\,{\rm
  because}\atop\eta(x_1)=\eta(x_2)=0}-
\int\limits_{x_1}^{x_2}\Bigg(\frac{d}{dx}\frac{\partial f}{\partial
  y'}\Bigg)\eta(x)\,dx\nonumber\\
&=&\int\limits_{x_1}^{x_2}\left(\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}\right)\eta(x)\,dx
\end{eqnarray}
To make $J$ extremal, $\partial J/\partial\alpha$ needs to vanish for all
deviations $\eta(x)$, and thus the expression in the parentheses needs to
vanish. As the expression has to be 
evaluated for $\alpha=0$, the expression in parentheses leads to a
differential equation for the {\em optimal} $y(x)$:
\begin{equation}\label{eq:eulerequation}
\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}=0
\end{equation}
This condition is called the {\em Euler equation}, and provides a differential
equation to find the function $y(x)$ that maximizes or minimizes $J$. This is
a purely mathematical result. For
$f=L(x,\dot x, t)$, (\ref{eq:eulerequation})
has exactly the form of the Lagrange equation (\ref{eq:LagrangeEoM}) derived
from Newton's laws earlier. Therefore, in a mechanics context, the equations
of motion of this form are also called {\em Euler-Lagrange equations}.

\subsubsection{Second from of the Euler equation}
The Euler equation (\ref{eq:eulerequation}) can be simplified if the function
$f$ does not explicitly depend on $x$:
\begin{equation}
f=f(y,y')
\end{equation}
To arrive there, one starts with the total differential
\begin{equation}\label{eq:somedifferential}
\frac{df}{dx}=\underbrace{\frac{\partial f}{\partial x}}_{=0}+\frac{\partial f}{\partial y}y'+\frac{\partial f}{\partial y'}y''\,.
\end{equation}
The first term vanishes because $f$ does not explicitly depend on $x$. By evaluating
\begin{equation}
\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}\right)=
y''\frac{\partial f}{\partial y'}+y'\frac{d}{dx}\frac{\partial f}{\partial y'}
\end{equation}
and substituting the first term on the right side with the last term in
(\ref{eq:somedifferential}), one finds
\begin{eqnarray}
\frac{d}{dx}\left(y'\frac{\partial f}{\partial y'}\right)&=&
\frac{df}{dx}-\frac{\partial f}{\partial
  y}y'+y'\frac{d}{dx}\frac{\partial f}{\partial y'}\nonumber\\
&=&\frac{df}{dx}+y'\underbrace{\left(\frac{d}{dx}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial y}\right)}_{=0}
\end{eqnarray}
The last term vanishes because of the Euler
equation (\ref{eq:eulerequation}). The rest can be written as 
\begin{equation}\label{eq:EulerSecondForm}
\frac{d}{dx}\left(f-y'\frac{\partial f}{\partial y'}\right)=0\quad{\rm
  or}\quad
f-y'\frac{\partial f}{\partial y'}=const.
\end{equation}
This is the so-called {\em second from} of the Euler equation for $y(x)$ to
minimize/maximize the functional $J$ in (\ref{eq:costfunctional}).
\subsubsection{Example: shape of a heavy rope}
As a simple example for a variational problem, we consider the shape of a
heavy chain or rope suspended between two poles:

\begin{figure}[h]
\centerline{\includegraphics{heavychain.eps}}
\end{figure}

The principle that defines the shape of the rope is the demand that the
potential energy of the chain at rest should be minimal, because all
deviations from  that configuration would drive the system into motion, which
would eventually be converted into heat via friction.
The problem is completely defined by specifying the length $l$ of the
rope, and the spacing $2a<l$ between the poles.

The total potential energy $U$ in the gravitational acceleration $g$ of the rope
with a line density (i.e., mass per length) $\rho$  is given by
\begin{equation}
U=\int\!\rho\,g\,y\,ds\,,
\end{equation}
where the integration is carried out along the rope, with a line element
$ds$. This integration along the rope can be expressed by an integration along
$x$ with the line element $ds=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}dx$:
\begin{equation}
\raisebox{-2em}{\includegraphics{lineelement.eps}}\qquad U=\rho g\int\limits_{-a}^ay\sqrt{1+y'^2}dx
\end{equation}
This is a variational problem of the ``second form''
(\ref{eq:EulerSecondForm}) with $f=y\sqrt{1+y'^2}$. With the constant $c$
required by the Euler equation in the second form, we get
\begin{eqnarray}
c=f-y'\frac{\partial f}{\partial y'}
&=&y\sqrt{1+y'^2}-y'\frac{yy'}{\sqrt{1+y'^2}}\nonumber\\
&=&\frac{y(1+y'^2)}{\sqrt{1+y'^2}}-\frac{yy'^2}{\sqrt{1+y'^2}}=\frac{y}{\sqrt{1+y'^2}}\,,
\end{eqnarray}
which can be further transformed into
\begin{equation}
  c\,y'=c\,\frac{dy}{dx}=\sqrt{y^2-c^2}\quad{\rm or}\quad \frac{c\,dy}{\sqrt{y^2-c^2}}=dx
\end{equation}
Integration on both sides leads to
\begin{equation}
{\rm arccosh}\, \frac{y}{c}=\frac{1}{c}(x+x_0)
\end{equation}
or finally
\begin{equation}
y(x)=c\,\cosh\left(\frac{x+x_0}{c}\right)\,,
\end{equation}
with the two integration constants $c$ and $x_0$. Since $\cosh
z=(e^z+e^{-z})/2$ is an even function, we choose $x_0=0$ in the middle of the
rope. To fix the final constant $c$ from the rope length, we need an
expression for the length, and with $y'=\sinh (x/c)$ we find
\begin{eqnarray}
l&=&\int\limits_{-a}^a\sqrt{1+y'^2}\,dx=\int\limits_{-a}^a\sqrt{1+\sinh^2\frac{x}{c}}\,dx=
\int\limits_{-a}^a\cosh\frac{x}{c}\,dx\nonumber\\
&=&c\sinh\frac{x}{c}\Bigg|_{-a}^a=2c\sinh\frac{a}{c}
\end{eqnarray}
This leads to a transcendental equation
\begin{equation}
  \sinh z=z\,\frac{l}{2a}\quad{\rm with}\quad z=\frac{a}{c}\,,
\end{equation}
which finally has to be solved numerically for $z$ and consequently $c$.

\subsubsection{Euler equations with several dependent variables}
The Lagrange equations (\ref{eq:LagrangeEoM}) typically involve several
dependent variables, i.e., all coordinates of all particles. This corresponds
to $N$ individual functions $y_k(x)$ that all depend on the same $x$.
The functional $f$ in (\ref{eq:costfunctional}), 
\begin{equation}
f=f\left(\{y_k(x), y'_k(x)\}; x\right)\,,
\end{equation}
depends now on an ensemble $\{y_k(x), y'_k(x)\}$ of variables 
$y_k$ and their derivatives $y'_k$ with respect to $x$ in the same way as in
(\ref{eq:newtonequations}). For the variational argument, one defines a variation for component $y_k$ in the same
way as in (\ref{eq:smalldeviation1}),
\begin{equation}
y_k(\alpha,x)=y_k(x)+\alpha\eta_k(x)\,.
\end{equation}
The requirement to the extremal condition on $J$ in (\ref{eq:costfunctional})
becomes 
\begin{equation}\label{eq:costfunctionmany}
\frac{\partial
  J}{\partial\alpha}=\int\limits_{x_k}^{x_2}\sum\limits_k\left(\frac{\partial
    f}{\partial y_k}-\frac{d}{dx}\frac{\partial f}{\partial y'_k}\right)\eta_k(x)\,dx\stackrel{!}{=}0\,.
\end{equation}
Since all deviation functions $\eta_k(x)$ are independent, all the
parentheses in the sum above have to vanish, which results in a system of
differential equations:
\begin{equation}\label{eq:eulerequations}
\frac{\partial
    f}{\partial y_k}-\frac{d}{dx}\frac{\partial f}{\partial y'_k}=0 \quad{\rm
    for\,\,all}\quad k=1\ldots N
\end{equation}
This is exactly of the form that allows deriving the Lagrange equations from
the Hamilton principle in (\ref{eq:hamiltonprinciple}), where $f$ becomes the
Lagrange function $L(\{x_k,\dot{x}_k\},t)$, and the independent variable $x$
is replaced with time $t$.


\subsection{Generalized coordinates}
The Hamilton principle states that a physical system takes the trajectory that
makes the variation of the action integral vanish:
\begin{equation}
\delta S=\delta\!\int\limits_{t_1}^{t_2}L\,dt=\delta\!\int\limits_{t_1}^{t_2}(T-U)\,dt=0\,.
\end{equation}
Earlier, the scalar Lagrange function $L$ was expressed in a set of
Cartesian coordinates, $\{x_k\}$, and a set of corresponding velocities,
$\{\dot{x}_k\}$, and 
eventually the time $t$ explicitly. Then, the Euler-Lagrange equations
(\ref{eq:eulerequations}) provide equations of motion for the whole physical
system:
\begin{equation}\label{eq:eulerlagrange}
\frac{\partial
    L}{\partial x_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_k}=0 \quad{\rm
    for\,\,all}\quad k=1\ldots N
\end{equation}
These equations of motion are equivalent to the equations of motion we derived
out of Newton's laws, so there would be no advantage in introducing a new
principle to derive the equations of motion of a system.

In the variational calculus argument, there was no explicit reference
that Cartesian coordinates needed to be used to write down the Lagrange
function. The Euler-Lagrange equations (\ref{eq:eulerlagrange}) were obtained
buy simply replacing the $y_k$ in 
(\ref{eq:eulerequations}) by coordinates $x_k$, and similarly their
derivatives with respect to the independent parameter, and switching to the
independent variable $t$. The Hamilton principle does, however, not state
anywhere that the Cartesian coordinates have to be used for describing the
system. Therefore, it is fine to use other coordinates to describe the system,
and the Hamilton principle provides a way to find the equations of motion in
other than Cartesian coordinate systems. This is an important advantage of the
Hamilton principle in comparison to the approach by using Newton's laws.

\subsubsection{Simple example: plane pendulum}\label{sec:pendulum}
To demonstrate this, we consider the relatively simple problem of a pendulum,
made up by a mass $m$ on a massless string of length $l$ in the gravitational
acceleration $\vec{g}$:
\begin{figure*}[h]
\centerline{\includegraphics{pendulum.eps}}
\end{figure*}

To come up with the equation of motion from Newton's laws, we first recognize
that the motion of the mass is constrained on a circular trajectory. An
adequate coordinate to express this is the angle $\theta$, with a fixed distance
$l$ to the origin. Forces acting on the particle are the force
$\vec{F}_g=m\vec{g}=-mg\,\vec{e}_y$ induced by the gravitational acceleration
$\vec{g}$, and a {\em constraining force } $\vec{F}_c$ exerted by the
string on the mass to keep it on a circular trajectory, which is aligned with
the string, 
\begin{equation}\label{eq:pendulumconstraint1}
\vec{F}_c=-F_c\,\vec{e}_r=-F_c\,(\vec{e_x}\sin\theta-\vec{e_y}\cos\theta).
\end{equation}
For a circular motion, $r=l=const.$, and therefore $\dot{r}=0$ and $\ddot{r}=0$. According to
(\ref{eq:radialmotoion2}), the projection $a_r=\ddot\vec{x}\cdot\vec{e}_r$ of the
total acceleration $\ddot\vec{x}$ in radial direction is  given by
$a_r=\ddot{r}-r\dot\theta^2=-l\dot\theta^2$. With the constraining force
pointing in inverse radial direction, $\vec{F}_c=-F_c\,\vec{e}_r$, one can
determine its amplitude $F_c$:
\begin{eqnarray}
(\vec{F}_g+\vec{F}_c)\cdot\vec{e}_r&=&(m\,\ddot\vec{x})\cdot\vec{e}_r\nonumber\\
-mg\,(\vec{e}_y\cdot\vec{e}_r)-F_c&=&m\,a_r\nonumber\\
-mg\,(-\cos\theta)-F_c&=&-m\,l\dot\theta^2\nonumber\\
F_c&=&m\,(g\cos\theta+\,l\dot\theta^2)\,.\label{eq:pendulumconstraint2}
\end{eqnarray}
With knowledge of $F_c$, we can finally
evaluate the total force, and use Newton's law to make the connection to the
acceleration $\ddot\vec{x}$, again using (\ref{eq:radialmotoion2}):
\begin{eqnarray}
m\,\ddot\vec{x}&=&\vec{F}_g+\vec{F}_c\nonumber\\
m\,(-l\dot\theta^2\vec{e}_r+l\ddot\theta\,\vec{e}_\theta)&=&
-mg\,\vec{e}_y-m\,(g\cos\theta+l\dot\theta^2)\,\vec{e}_r
\end{eqnarray}
The terms with $l\dot\theta^2$ cancel, and by multiplying the last equation with
$\vec{e}_\theta=-\vec{e}_x\cos\theta+\vec{e}_y\sin\theta$, dividing by $m$ and
using $\vec{e}_r\cdot\vec{e}_\theta=0$, we get
\begin{equation}
l\ddot\theta=-g\,(\vec{e}_y\cdot\vec{e}_\theta)=-g\sin\theta
\end{equation}
or finally
\begin{equation}\label{eq:eqmotionpendulum1}
\ddot\theta+\frac{g}{l}\sin\theta=0\,.
\end{equation}
Steps (\ref{eq:pendulumconstraint1}) and (\ref{eq:pendulumconstraint2}) may
seem unnecessarily complicated, because the constraining force was explicitly
calculated, but this would be the procedure according to Newton's laws, taking
care of all forces acting on the mass. One could have cut some
corners by only looking at the projection of all forces on $\vec{e}_\theta$,
and notice that the only force that contributes here would be $\vec{F}_g$,
while the constraining force is orthogonal to $\vec{e}_\theta$, and would not
need to be evaluated explicitly.

Now, we compare the strategy to obtain the equation of motion with the one
provided by the Lagrange formalism. We start out with writing down the kinetic
energy,
\begin{equation}
T=\frac{1}{2}mv^2=\frac{1}{2}ml^2\dot\theta^2\,,
\end{equation}
and the potential energy of a mass in gravitational acceleration,
\begin{equation}
U=mgh-U_0=mgl\,(1-\cos\theta)-U_0=-mgl\cos\theta\,,
\end{equation}
where we could subtract the constant offset $U_0=mgl$ because this does not
affect the dynamics of the system. The Lagrange function then can be written as 
\begin{equation}\label{eq:lagrangependulum}
L=T-U=\frac{1}{2}ml^2\dot\theta^2+mgl\cos\theta\,,
\end{equation}
which is now only a function of the dynamic variables $\theta$ and
$\dot\theta$. Since Hamilton's principle makes no statement what coordinates
can be used, as long as one can define a meaningful potential and kinetic
energy, one could just use (\ref{eq:eulerlagrange}) for the single variable
$\theta$, and get the Euler-Lagrange equations of motion:
\begin{equation}
\frac{\partial L(\theta,\dot\theta)}{\partial
  \theta}-\frac{d}{dt}\frac{\partial L(\theta,\dot\theta)}{\partial
  \dot\theta}=0\,.
\end{equation}
By carrying out the differentiations of $L$, one finds
\begin{equation}
-mgl\sin\theta-\frac{d}{dt}ml^2\dot\theta=-mgl\sin\theta-ml^2\ddot\theta=0\,.
\end{equation}
Dividing the last part by $-ml^2$, we reproduce the equation of
motion (\ref{eq:eqmotionpendulum1}):
\begin{equation}\label{eq:eqmotionpendulum2}
\ddot\theta+\frac{g}{l}\sin\theta=0\,.
\end{equation}
In this derivation, there was no need to work out any forces. A lucky choice
of the right coordinate, $\theta$ in this case, took implicitly care of
the constraint that the mass has to move on a circle.

\subsubsection{Generalized coordinates and velocities}
The treatment of the pendulum above is only an example of describing a system
by {\em generalized coordinates}, which are usually referred to as a set
$\{q_k\}$. Together with the set of their 
corresponding temporal derivatives, $\{\dot{q}_k\}$, referred to as {\em
  generalized velocities}, they give a complete
description of the state of a physical system at any point in time. Again, the
system can be made up by a number of particles. Then, Hamilton's principle
leads directly to a set of equations of motion for the whole system:
\begin{equation}\label{eq:generaleulerlagrange}
\frac{\partial
    L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_k}=0 \quad{\rm
    for\,\,all}\quad k=1\ldots N
\end{equation}

A few remarks on generalized coordinates seem to be in order:
\begin{itemize}
\item Generalized coordinates do not need to have the dimension of
  a length, like Cartesian coordinates do by specifying the respective
  distance from a reference point to the position of a mass point.
\item Likewise, generalized velocities do not need to have the dimension of a
  length per time, the dimension of the traditional velocity.
\item Generalized coordinates of a particle can be functions of various
  coordinates of one particle, but can also be combinations of coordinates
  from different particles.
\item To some extent, generalized coordinates may explicitly contain the
  time.
\item Generalized coordinates may be used to take care of constraints of a
  system, like in the pendulum example. A careful analysis of the degrees of
  freedom of a system can eliminate the explicit need to consider constraints.
\end{itemize}
The only thing that is really required for generalized coordinates to be
useful for coming up with the equations of motion of a physical system is that
all the degrees of freedom of the system are properly captured, and that it is
possible to formulate a Lagrange function that only depends on the $\{q_k\}$
and $\{\dot{q}_k\}$.

\subsubsection{Point transformations between traditional and generalized coordinates}
One can make the transition between the traditional Cartesian coordinates
$\{x_k\}$ and the generalized coordinates $\{q_k\}$ of a system slightly more
explicit by introducing a specific transformation between the two sets, and
then showing that the Euler-Lagrange equations in the generalized coordinates
follow from the Euler-Lagrange equations in the original coordinates. This
step, however, is not necessary to see that (\ref{eq:generaleulerlagrange})
can describe the complete dynamics of a system, but it illustrates what
transformations can be used to construct generalized coordinates.

A property for a set of generalized coordinates $\{\dot{q}_k\}$ is that
all the degrees of freedom of the system are captured. Then, there must exist
a transformation that converts between the old and new set of coordinates, so
the old coordinates may be expressed as a function of the new coordinates, and
possible the time $t$:
\begin{equation}\label{eq:pointtransf1}
x_k=x_k(\{q_l\},t)
\end{equation}
This notation means that $x_k$ is an explicit function of all the $q_l$,
and the time $t$. Then, the temporal derivative of $q_k$ can be calculated:
\begin{equation}\label{eq:pointtransf2}
\dot{x}_k=\frac{\partial x_k}{\partial t}+\sum\limits_l\frac{\partial
  x_k}{\partial q_l}\frac{dq_l}{dt}= 
\frac{\partial x_k}{\partial t}+\sum\limits_l\frac{\partial
  x_k}{\partial q_l}\,\dot{q}_l
\end{equation}
This makes $\dot{x}_k$ an explicit function of the set $\{q_l\}$ (via the
derivatives of $x_k$ with respect to $q_l$), the set $\{\dot{q}_l\}$, and the
time. The Lagrange function in the old coordinates, $L$, and the one in new
coordinates, $L'$, should be the same, but of course the functions $L$ and
$L'$ take a different form in the respective coordinate sets -- this is why
there is a different symbol, $L'$:
\begin{equation}
  L(\{x_k\},\{\dot{x}_k\},t)=L(\{x_k(\{q_l\},t)\}, \{\dot{x}_k(\{q_l\},
  \{\dot{q}_l\},t)\}=L'(\{q_l\},\{\dot{q}_l\},t)
\end{equation}
We now evaluate the derivatives we need for the Euler-Lagrange equations in
the new coordinates:
\begin{equation}
\frac{\partial L'}{\partial q_l}=\sum\limits_k\frac{\partial L}{\partial
  x_k}\frac{\partial x_k}{\partial  q_l}+\frac{\partial L}{\partial
  \dot{x}_k}\frac{\partial \dot{x}_k}{\partial  q_l}
\end{equation}
and 
\begin{equation}
\frac{\partial L'}{\partial \dot{q}_l}=\sum\limits_k\Bigg(
\frac{\partial L}{\partial
  x_k}\underbrace{\frac{\partial x_k}{\partial \dot{q}_l}}_{=0}+\frac{\partial L}{\partial
  \dot{x}_k}\underbrace{\frac{\partial \dot{x}_k}{\partial
    \dot{q}_l}}_{=\partial x_k/\partial q_l}\!\!\Bigg)=
\sum\limits_k\left(\frac{\partial L}{\partial \dot{x}_k}
  \frac{\partial x_k}{\partial q_l}\right)
\end{equation}
The first term vanishes because in (\ref{eq:pointtransf1}), $x_k$ does not
explicitly depend on the velocity $\dot{q}_l$. The change in the second
derivative can be seen from (\ref{eq:pointtransf2}) because the only term in
$\dot{x}_k$ that depends on $\dot{q}_l$ is the one with $\partial x_k/\partial
q_l$ (don't get confused with the same $l$ index -- in
(\ref{eq:pointtransf2}), this is a summation index, and can be replaced
with, say, $m$. Then, $\partial \dot{q}_{m}/\partial q_l=\delta_{lm}$ and
the sum vanishes...).

Then, the ``new'' Euler-Lagrange equation can then simply be calculated:
\begin{eqnarray}\label{eq:newelag}
  \frac{\partial L'}{\partial q_l}-\frac{d}{dt}\frac{\partial L'}{\partial
    \dot{q}_l} &=&
  \sum\limits_k\left(\frac{\partial L}{\partial
      x_k}\frac{\partial x_k}{\partial  q_l}+\frac{\partial L}{\partial
      \dot{x}_k}\frac{\partial \dot{x}_k}{\partial  q_l}\right)
  -\frac{d}{dt}\left(
    \sum\limits_k\frac{\partial L}{\partial \dot{x}_k}
    \frac{\partial x_k}{\partial q_l}\right)\nonumber\\
  &=&\sum\limits_k\left[
    \frac{\partial x_k}{\partial  q_l}\Bigg(
      \underbrace{\frac{\partial L}{\partial x_k}-
    \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_k}}_{=0}\Bigg)
  +\frac{\partial L}{\partial \dot{x}_k}\left(
    \frac{\partial \dot{x}_k}{\partial q_l}
    -\frac{d}{dt}\frac{\partial x_k}{\partial q_l}\right)
\right]
\end{eqnarray}
The first parenthesis vanishes because of the Euler-Lagrange equations in the
original coordinates $\{x_k\}$. For the second term, one explicitly calculates 
\begin{eqnarray}
  \frac{d}{dt}\frac{\partial x_k}{\partial q_l}&=&
  \frac{\partial^2x_k}{\partial q_l\partial t}+
  \sum\limits_m\frac{\partial^2x_k}{\partial q_l\partial
    q_m}\dot{q}_m\nonumber\\
  &=&\frac{\partial}{\partial q_l}\left(\frac{\partial x_k}{\partial t}+
\sum\limits_m\frac{\partial x_k}{\partial q_m} \dot{q}_m\right)
=\frac{\partial}{\partial
  q_l}\frac{dx_k}{dt}=\frac{\partial\dot{x}_k}{\partial q_l}
\end{eqnarray}
With this, also the second term on the right side in (\ref{eq:newelag})
vanishes, and the Euler-Lagrange equations in the new coordinates $\{q_l\}$
and the corresponding velocities are recovered. 

\subsubsection{Generalized momenta and cyclic coordinates}
Newton's laws made an explicit statement on the quantity momentum associated
with a Cartesian coordinate $x_k$, defined in (\ref{eq:momentumfromT}) as
$p_k=m_k\dot{x}_k$. To come to a similarly useful definition for generalized
coordinates, we first try to extract the linear momentum $p_k$ from the
Lagrange function. We first note that the required velocity $\dot{x}_k$
appears in the expression (\ref{eq:lagKineticenergy}) of the kinetic energy,
\begin{equation}
T=\sum\limits_{k}\frac{1}{2}m_{k}\dot{x}_{k}^2\,.
\end{equation}
The momentum $p_k$ can be extracted from $T$ by differentiating with respect
to $\dot{x}_k$:
\begin{equation}
\frac{\partial T}{\partial\dot{x}_k}=m_k\dot{x}_k=p_k
\end{equation}
Since Lagrange function is the difference between $T$ and a potential $U$
that should not depend on the velocities, one can obtain the momentum directly
by differentiating the Lagrange function with respect to $\dot{x}_k$. This can
be used to define a {\em generalized momentum}, if the Lagrange function is
expressed in generalized coordinates $\{q_l\}$ and  generalized velocities
$\{\dot{q}_l\}$:
\begin{equation}\label{eq:momentumdef2}
p_k:=\frac{\partial L}{\partial \dot{q}_k}
\end{equation}

The difference between generalized momentum and the traditional momentum can
e.g. be seen with the Lagrange function of a charged particle in an
electromagnetic field from (\ref{eq:lagrangianchargedparticle}): 
\begin{equation}
L=\sum\limits_{i=1}^3\frac{1}{2}m\dot{x}_i^2-q\Phi(\vec{x})+\sum\limits_{i=1}^3q\dot{x}_iA_i(\vec{x})\,.
\end{equation}
The corresponding generalized momentum to the coordinate $x_i$ is
\begin{equation}
p_i=\frac{\partial L}{\partial \dot{x}_i}=m\dot{x}_i+qA_i(\vec{x})\,,
\end{equation}
which contains not only the kinetic momentum $m\dot{x}_i$ of the particle's
motion, but also a contribution $qA_i$ from the vector potential.

The generalized momentum can also be used to rewrite the Euler-Lagrange
equation:
\begin{equation}
\frac{\partial L}{\partial q_k} -\frac{d}{dt}\frac{\partial L}{\partial
  \dot{q}_k} = \frac{\partial L}{\partial q_k}-\frac{d}{dt}p_k=0\quad{\rm
  or}\quad \dot{p}_k=\frac{\partial L}{\partial q_k}
\end{equation}
This has an important consequence if the Lagrange function does not depend on
a particular generalized coordinate $q_k$. Such coordinates are called {\em
  cyclic}. If $q_k$ is a cyclic coordinate, $\partial L/\partial q_k=0$, and
\begin{equation}\label{eq:momentumconservation}
\dot{p}_k=0\quad{\rm or}\quad p_k=const.
\end{equation}
So if $L$ does not depend on a coordinate $q_k$, and is therefore symmetric
with respect to a translation in $q_k$, the corresponding generalized momentum
does not change in time or is {\em conserved}.

We will expand more on this important connection between symmetries and
conserved quantities that holds for any generalized coordinate and its
canonically conjugated 
momentum, because it will help to explore symmetries in problems.

\subsection{Noether's theorem: Symmetries and conservation laws}
The conservation of the canonically conjugated momentum of a cyclic
generalized coordinate deserves slightly more attention, because it can be
expanded to different symmetries. The mathematically rigorous version of this
idea was formulated by Emmy
Noether\footnote{\href{https://en.wikipedia.org/wiki/Emmy_Noether}{A. Emmy
    Noether}, 1882-1935}, and the connection between symmetries and 
conserved quantities is known as the {\em
  \href{https://en.wikipedia.org/wiki/Noether's_theorem}{Noether theorem}}. 
In a simple form of this theorem, symmetries are expressed as an invariance of
the Lagrange function against small symmetry transformations. Here, we
consider three different symmetry examples: translation symmetry in space,
rotation symmetry, and translation symmetry in time. However, this is a very
general principle that reaches way beyond classical mechanics, and is
e.g. heavily used in elementary particle physics.

\subsubsection{Translation symmetry and momentum conservation} 
%We have seen a translational symmetry of a physical system already in the form
%of cyclical coordinates:
If a Lagrangian does not explicitly depend on a particular coordinate $q_k$,
i.e., $q_k$ is cyclic, $L$ is symmetric with respect to a
translation in this coordinate; this caused momentum
conservation (\ref{eq:momentumconservation}).

We now use a slightly different way of noting this symmetry that will
allow treating the rotational symmetry easier later. First, we consider a
transformation 
\begin{equation}
\vec{q}\rightarrow\vec{q'}=\vec{q}+\delta\vec{q}\,,
\end{equation}
where $\vec{q}$ is a vector of (generalized) coordinates, and $\delta\vec{q}$
a small displacement vector in a particular direction. We can decompose this
displacement with the help of a unit vector set $\{\vec{e}_l\}$:
\begin{equation}
\delta\vec{q}=\sum\limits_l\delta q_l\,\vec{e}_l\,,
\end{equation}
with small time-independent displacements $\delta q_l$. Then, the change
$\delta L$ of the Lagrange function under such a transformation is given by
\begin{equation}
\delta L=\sum\limits_l\left(\frac{\partial L}{\partial q_l}\delta
  q_l+\frac{\partial L}{\partial\dot{q}_l}\delta\dot{q}_l\right)\,.
\end{equation}
The last term vanishes, as $\delta\dot{q}_l=d(\delta q_l)/dt$ and the small
displacements $\delta_l$ do not 
depend on time. A symmetry with respect to a small translation means that the
Lagrange function should not change under this transformation, or $\delta
L=0$. Thus,
\begin{equation}
\delta L=\sum\limits_l\frac{\partial L}{\partial q_l}\delta q_l=
\sum\limits_l\dot{p}_l\delta q_l=\dot\vec{p}\cdot \delta\vec{q}=0\,.
\end{equation}
We can interpret this result in the following way: if the system (and
therefore the Lagrangian) 
has a translational symmetry in the direction $\delta\vec{q}$, the
{\em projection} of the generalized momentum $\vec{p}$ on this direction does
not change in time, or is {\em conserved}.

\subsubsection{Rotational symmetry and angular momentum conservation}
\label{ssec:noether-rotation}
We recall from section \ref{ssec:smallrotations} that infinitesimal
rotations can be described by a vector $\delta\vec{\theta}$, which according to
(\ref{eq:smallmove}) causes a displacement of something at position $\vec{r}$
(referenced to a point on the rotation axis) by
\begin{equation}
\delta\vec{r}=\delta\vec{\theta}\times\vec{r}\,,\quad{\rm and\,\,also}\quad
\delta\dot\vec{r}=\delta\vec{\theta}\times\dot\vec{r}
\end{equation}
because the vector $\delta\vec{\theta}$ should not
depend on time. Under this position transformation
\begin{equation}
\vec{r}\rightarrow\vec{r}'=\vec{r}+\delta\vec{r}\,,
\end{equation}
the change of the Lagrange function is given by
\begin{eqnarray}
\delta L&=&\sum\limits_{l=1}^3\Bigg(\underbrace{\frac{\partial L}{\partial x_l}}_{=\dot{p}_l}\delta x_l+
\underbrace{\frac{\partial L}{\partial \dot{x}_l}}_{=p_l}\delta
\dot{x}_l\Bigg)\nonumber\\
&=&\sum\limits_{l=1}^3\dot{p}_l\,\delta x_l+p_l\,\delta\dot{x}_l\nonumber\\
&=&\dot\vec{p}\cdot\delta\vec{r}+\vec{p}\cdot\delta\dot\vec{r}
=\dot\vec{p}\cdot(\delta\vec{\theta}\times\vec{r})+\vec{p}\cdot(\delta\vec{\theta}\times\dot\vec{r})\,.
\end{eqnarray}
With the cyclic permutation symmetry of the triple product\footnote{The
  product $\vec{a}\cdot(\vec{b}\times\vec{c})$ is used to calculate the volume
  of a \href{https://en.wikipedia.org/wiki/Parallelepiped}{parallelepiped} spanned by the three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.}  
$\vec{a}\cdot(\vec{b}\times\vec{c})= \vec{b}\cdot(\vec{c}\times\vec{a})$,
\begin{eqnarray}
\delta
L&=&\delta\vec{\theta}\cdot(\vec{r}\times\dot\vec{p})+\delta\vec{\theta}\cdot(\dot\vec{r}\times\vec{p})\nonumber\\
&=&\delta\vec{\theta}\cdot(\vec{r}\times\dot\vec{p}+\dot\vec{r}\times\vec{p})\nonumber\\
&=&\delta\vec{\theta}\cdot\left(\frac{d}{dt}(\vec{r}\times\vec{p})\right)\nonumber\\
&=&\delta\vec{\theta}\cdot\dot\vec{L}\,,
\end{eqnarray}
where $\vec{L}$ was the angular momentum vector introduced in
(\ref{eq:angularmomentumdef}). If the physical system is symmetric with respect
to a rotation defined by $\delta\vec{\theta}$, the Lagrangian does not change,
i.e., $\delta L$=0. Then, 
\begin{equation}
\delta\vec{\theta}\cdot\dot\vec{L}=0\,,
\end{equation}
which means that the projection of the angular momentum vector $\vec{L}$ on the
axis of rotational symmetry, defined by the direction of $\vec{\theta}$, is a
constant in time, or is conserved. An example would be a spherical pendulum,
where a mass $m$ suspended by a string of length $l$ in gravity is allowed
oscillating freely. We leave it to the reader to work out the Lagrange function
for that case, but it should be obvious that neither the potential nor the
kinetic energy depend on the polar angle, thus the angular momentum component
along the vertical direction is conserved.

For a spherically symmetric problem, the Lagrangian is independent of
rotations around all directions in space, thus, all components of the angular
momentum are conserved.

\subsubsection{Invariance against translation in time and energy conservation}
\label{ssec:Noether3}
We now consider a different symmetry, where the physical system is invariant
by a translation in time:
\begin{equation}
t\rightarrow t'=t+\delta t\,.
\end{equation}
Then, the change of the Lagrangian under small changes $\delta t$ must vanish,
or
\begin{equation}
\frac{\partial L}{\partial t}=0\,,
\end{equation}
i.e., the Lagrange function is not explicitly time dependent.  Under this
condition, we can calculate the total differential of the Lagrange function
with respect to time (because the system can still evolve, causing $L$ to
change over time):
\begin{eqnarray}\label{eq:totaldifflagrangian}
\frac{d}{dt}L&=&\sum\limits_k\left(\frac{\partial L}{\partial q_k}\dot{q}_k+
\frac{\partial L}{\partial \dot{q}_k}\ddot{q}_k\right)\nonumber\\
&=&\sum\limits_k\left(\left[\frac{d}{dt}\left(\frac{\partial L}{\partial
      \dot{q}_k}\right)\right]\dot{q}_k+\frac{\partial L}{\partial
    \dot{q}_k}\ddot{q}_k\right)\nonumber\\
&=&\sum\limits_k\frac{d}{dt}\left(\frac{\partial L}{\partial
      \dot{q}_k}\dot{q}_k\right)
\end{eqnarray}
From the first line to the second, the Lagrange equation (\ref{eq:LagrangeEoM})
allows replacing $\partial L/\partial q_k$, and from the second to third
line, the product rule for differentiations was applied. The differentiation
with 
respect to time in the last line can be taken out of the sum, and the whole
equation (\ref{eq:totaldifflagrangian}) can be written as 
\begin{equation}\label{eq:constanthamiltonian}
\frac{d}{dt}\left(\sum\limits_k\frac{\partial L}{\partial \dot{q}_k}\dot{q}_k-L\right)=0\quad{\rm or}\quad\frac{d}{dt}H=0
\end{equation}
The quantity in the parenthesis above is defined as the {\em Hamilton
  function}
\begin{equation}\label{eq:hamiltoniandefinition}
  H:=\sum\limits_k\frac{\partial L}{\partial \dot{q}_k}\dot{q}_k-L
\end{equation}
of the problem, and is constant in time, or conserved if the
physical system and therefore the Lagrangian is symmetric with respect to a
translation in time. In many cases, the Hamilton function
is the same as the total energy. In those cases, the invariance of $L$ under
translations in time means that the total energy of the system is conserved,
as we have seen in section \ref{ssec:energyconservation}
and \ref{ssec:energyconservationensemble}. Equation (\ref{eq:constanthamiltonian})
expresses energy conservation for a physical system described in {\em
  generalized coordinates}, but with the same assumption on the
time-independence of $U$ as earlier. To identify $H$ with the total energy
$E$, there is another constraint to consider in how the generalized
coordinates are connected with the Cartesian coordinates. 

\subsection{Hamilton function and total energy}\label{ssec:HamiltonTotalenergy}
To interpret the Hamilton function defined in
(\ref{eq:hamiltoniandefinition}),  we  detour to an aspect of the
total kinetic energy $T$ of the system. In Cartesian coordinates, the kinetic
energy can be written as
\begin{equation}
T=\sum\limits_k\frac{1}{2}m_k\dot{x}_k^2\,.
\end{equation}
Via the point transformations (\ref{eq:pointtransf1}) between $\{x_k\}$ and
$\{q_l\}$, and using (\ref{eq:pointtransf2}), the kinetic energy can be
expressed via generalized velocities $\dot{q}_l$:
\begin{eqnarray}\label{eq:kineticenergy}
T&=&\sum\limits_k\frac{1}{2}m_k
\left(\sum\limits_l\frac{\partial x_k}{\partial q_l}\dot{q}_l+\frac{\partial
    x_k}{\partial t}\right)
\left(\sum\limits_{l'}\frac{\partial x_k}{\partial q_{l'}}\dot{q}_{l'}+\frac{\partial x_k}{\partial t}\right)\nonumber\\
&=&\sum\limits_{l,l'}\left[
  \sum\limits_k\frac{1}{2}m_k\frac{\partial x_k}{\partial q_l}
  \frac{\partial x_k}{\partial q_{l'}}\right]
\dot{q}_l\dot{q}_{l'}+
\sum\limits_{l}\left[
  2\sum\limits_{k}\frac{1}{2}m_k\frac{\partial x_k}{\partial q_{l}}
                  \frac{\partial x_k}{\partial t}\right]\dot{q}_l+\left[
\sum\limits_{k}\frac{1}{2}m_k\left(\frac{\partial x_k}{\partial t}\right)^2\right]
\nonumber\\
&=&\sum\limits_{l,l'}a_{l,l'}\,\dot{q}_l\dot{q}_{l'}\,+\,
\sum\limits_{l}b_l\,\dot{q}_l\,+\,c\,,
\end{eqnarray}
where $a_{l,l'}$, $b_l$, and $c$ summarize the corresponding brackets in
the line before.
For time-independent point transformations from $\{x_k\}$ 
to generalized coordinates $\{q_l\}$, 
\begin{equation}
\frac{\partial x_k}{\partial t}=0\,,
\end{equation}
so  $b_l=c=0$ in the kinetic energy
expression (\ref{eq:kineticenergy}), which then simplifies to
\begin{equation}\label{eq:kineticenergy2}
T=\sum\limits_{l,l'}a_{l,l'}\,\dot{q}_l\dot{q}_{l'}\,.
\end{equation}
Then, the partial derivative of $T$ with respect to velocities $\dot{q}_k$ are
\begin{equation}\label{eq:twotee}
\frac{\partial T}{\partial\dot{q}_k}=\sum\limits_la_{l,k}\,\dot{q}_l+\sum\limits_{l'}a_{k,l'}\,\dot{q}_{l'}=\sum\limits_l(a_{l,k}+a_{k,l})\,\dot{q}_l
\end{equation}
In the first step, the two terms appear because in the double sum in
(\ref{eq:kineticenergy2}), $\dot{q}_k$ appears twice. In the second step, a
variable change $l'\rightarrow l$ for the sum allowed taking everything under
one sum. With this, one can evaluate the following sum:
\begin{eqnarray}\label{eq:homfunction1}
\sum\limits_k\dot{q}_k\frac{\partial T}{\partial\dot{q}_k}&=&
\sum\limits_{k,l}(a_{l,k}+a_{k,l})\,\dot{q}_k\dot{q}_l\nonumber\\
&=&2\sum\limits_{k,l}a_{l,k}\,\dot{q}_k\dot{q}_l=2T\,.
\end{eqnarray}
The step from the first to the second line simply involved a summation index
exchange $k\leftrightarrow l$ for one of the sums.

The relation (\ref{eq:homfunction1}) is special case of a property of
so-called \href{https://en.wikipedia.org/wiki/Homogeneous_function}{\em
  homogenous functions}. A multi-variable function $f(\{q_k\})$ is called
homogenous of degree $p$, if 
\begin{equation}
f(\{\lambda q_k\})=\lambda^pf(\{ q_k\})\,.
\end{equation} 
This means that if all parameters of $f$ are multiplied by a constant
$\lambda$, the value of $f$ is the value of the original function, multiplied
with $\lambda^p$. For example, the function $f(x,y)=x^2+y^2$ is homogenous of
degree $p=2$ in $x,y$. For homogenous functions, Euler\footnote{\href{https://en.wikipedia.org/wiki/Leonhard_Euler}{Leonhard Euler}, 1707-1783} showed that
\begin{equation}
\sum\limits_kq_k\frac{\partial f}{\partial q_k}=pf\,,
\end{equation}
which is referred to as {\em Euler's theorem on homogenous functions}.

After this small detour, we come back to the definition of the Hamilton
function (\ref{eq:hamiltoniandefinition}).
% \begin{equation}
% H=\sum\limits_k\frac{\partial L}{\partial \dot{q}_k}\dot{q}_k-L
% \end{equation}
If we assume that the potential $U$ in $L=T-U$ is independent of the velocities
$\dot{q}_k$, then 
\begin{equation}
\frac{\partial L}{\partial \dot{q}_k}=\frac{\partial T}{\partial \dot{q}_k}\,,
\end{equation}
and we can write
\begin{eqnarray}
H&=&\sum\limits_k\frac{\partial L}{\partial \dot{q}_k}\dot{q}_k-L\nonumber\\
&=&\sum\limits_k\frac{\partial T}{\partial \dot{q}_k}\dot{q}_k-L
=2T-L=2T-(T-U)=T+U
\end{eqnarray}
So the Hamilton function becomes the total energy (i.e., the sum of total
kinetic and potential energy) {\em under the assumptions we made} when
interpreting 
$H$. Explicitly, these assumptions were
\begin{itemize}
\item[(a)]the transformations form Cartesian to generalized coordinates
  $\{q_k\}$ used as parameters for $L$ and $H$ are time-independent, and 
\item[(b)]the potential energy $U$ is independent of the velocities $\dot{q}_k$.
\end{itemize}
An example where $H\neq E$ can be found with rotating coordinate systems,
since this makes the coordinate transformations to a inertial reference frame
time-dependent. Still, the total energy may be conserved.


\subsection{Non-conservative forces in the Lagrange formalism}\label{ssec:nonconservativeforces}
So far, the description of physical systems by the Euler-Lagrange formalism
required that forces can be expressed as a gradient of a potential, and do not
depend velocities, with notable exceptions like charged particles in
electromagnetic fields in section 
\ref{ssec:nonconservative}. However, friction forces and other dissipative
forces that can not be expressed by a potential or a Lagrange function are an
important part of many practical systems, so a mechanism to include these
phenomena in the very flexible Lagrange formalism is desirable.

To do this, we recall the definition of the work integral
(\ref{eq:ensemblework}) in section \ref{ssec:totalkineticenergy}, where the work
on an ensemble of particles transitioning from state $a$ to $b$ was defined
as  
\begin{equation}
W_{ab}=\sum\limits_\alpha\int\limits_a^b\vec{F}_\alpha\cdot d\vec{x}_\alpha\,.
\end{equation}
The position $\vec{x}_\alpha$ of
particle $\alpha$ can be expressed by generalized coordinates,
\begin{equation}
\vec{x}_\alpha=\vec{x}_\alpha(\{q_k\})\,,
\end{equation}
as long as the transformation between real and generalized
coordinates is not explicitly time dependent. The differential displacement
$d\vec{x}_\alpha$ can be written
\begin{equation}\label{eq:realdisplacement}
d\vec{x}_\alpha=\sum\limits_k\frac{\partial \vec{x}_\alpha}{\partial q_k}\,dq_k\,,
\end{equation}
and the differential $dW$ for the total work on the system by the forces
$\vec{F}_\alpha$ is
\begin{eqnarray}\label{eq:workwithgeneralforce}
dW&=&\sum\limits_\alpha\vec{F}_\alpha\cdot d\vec{x}_\alpha\nonumber\\
&=&\sum\limits_{\alpha,k}\vec{F}_\alpha\cdot\left(\frac{\partial
    \vec{x}_\alpha}{\partial q_k}\right)\,dq_k\nonumber\\
&=&\sum\limits_k\phi_k\,dq_k\,,
\end{eqnarray}
with the definition of the so-called {\em generalized force} 
\begin{equation}\label{eq:generalizedforcedefinition}
\phi_k:=\sum\limits_\alpha\vec{F}_\alpha\cdot\frac{\partial \vec{x}_\alpha}{\partial q_k}\,.
\end{equation}
Expression (\ref{eq:workwithgeneralforce}) holds for any type of force, and we
now try to reproduce the Lagrange equations. For that, we start with the
expression for the total kinetic energy as expressed by Cartesian
velocities,
\begin{equation}
T=\sum\limits_\alpha\frac{1}{2}m_\alpha(\dot\vec{x}_\alpha)^2\,,
\end{equation}
and calculate their derivatives with respect to generalized coordinates
$q_k$:
\begin{equation}
\frac{\partial T}{\partial q_k}=\sum\limits_\alpha
m_\alpha\,\dot\vec{x}_\alpha\cdot\frac{\partial 
  \dot\vec{x}_\alpha}{\partial q_k}
\end{equation}
From (\ref{eq:realdisplacement}), one finds
\begin{equation}
\dot\vec{x}_\alpha=\sum\limits_k\frac{\partial
  \vec{x}_\alpha}{\partial q_k}\,\dot{q}_k\quad{\rm and \,\,thus}\quad
\frac{\partial\dot\vec{x}_\alpha}{\partial\dot{q}_k}
=\frac{\partial\vec{x}_\alpha}{\partial q_k}
\end{equation}
With this, the derivative of the kinetic energy with respect to the
generalized velocity $\dot{q}_k$ is
\begin{equation}
\frac{\partial T}{\partial \dot{q}_k}=
\sum\limits_\alpha
m_\alpha\,\dot\vec{x}_\alpha\cdot \frac{\partial\dot\vec{x}_\alpha}{\dot{q}_k} = 
\sum\limits_\alpha
m_\alpha\,\dot\vec{x}_\alpha\cdot \frac{\partial\vec{x}_\alpha}{\partial q_k}\,.
\end{equation}
This allows assembling the kinetic energy dependent part of a Lagrange equation:
\begin{eqnarray}
\frac{\partial T}{\partial q_k}-\frac{d}{dt}\frac{\partial
  T}{\partial\dot{q}_k} &=& 
\sum\limits_\alpha
m_\alpha\,\dot\vec{x}_\alpha\cdot\frac{\partial\dot\vec{x}_\alpha}{\partial
  q_k}
-\frac{d}{dt}\left(\sum\limits_\alpha
m_\alpha\,\dot\vec{x}_\alpha\cdot\frac{\partial\vec{x}_\alpha}{\partial
  q_k}\right)\nonumber\\
&=&\sum\limits_\alpha m_\alpha\left[
\dot\vec{x}_\alpha\cdot\frac{\partial\dot\vec{x}_\alpha}{\partial q_k}
- \ddot\vec{x}_\alpha\cdot\frac{\partial\vec{x}_\alpha}{\partial q_k}
- \dot\vec{x}_\alpha\cdot\frac{d}{dt}\frac{\partial\vec{x}_\alpha}{\partial q_k}
\right]\nonumber\\
&=&-\sum\limits_\alpha m_\alpha\ddot\vec{x}_\alpha\cdot\frac{\partial\vec{x}_\alpha}{\partial q_k}\,.
\end{eqnarray}
Using Newton's equation of motion $\vec{F}_\alpha=m_\alpha\ddot\vec{x}_\alpha$
and the definition (\ref{eq:generalizedforcedefinition}) for the generalized
force, one obtains
\begin{equation}\label{eq:almostlagrange1}
\frac{\partial T}{\partial q_k}-\frac{d}{dt}\frac{\partial
  T}{\partial\dot{q}_k} =
-\sum\limits_\alpha\vec{F}_\alpha\cdot\frac{\partial\vec{x}_\alpha}{\partial
  q_k}=-\phi_k\,.
\end{equation}
The generalized force can always be split up in a part that can be derived
from a potential $U$, and a residual part $\phi'_k$,
\begin{equation}
\phi_k=-\frac{\partial U}{\partial q_k}+\phi'_k\,,
\end{equation}
with a velocity-independent potential $U=U(\{q_k\})$. Then, the equation of
motion (\ref{eq:almostlagrange1}) can be written as
\begin{equation}\label{eq:lagrangeEOMwithfriction}
\frac{\partial (T-U)}{\partial q_k}-\frac{d}{dt}\frac{\partial
  (T-U)}{\partial\dot{q}_k}=\frac{\partial L}{\partial q_k}
-\frac{d}{dt}\frac{\partial L}{\partial\dot{q}_k}=-\phi'_k\,,
\end{equation}
where $L=T-U$ is the usual Lagrange function that contains forces that can
be derived from a potential. Expression (\ref{eq:lagrangeEOMwithfriction})
resembles almost the set of Lagrange equations
(\ref{eq:generaleulerlagrange}), but this time, non-conservative forces like 
the friction forces  
seen in section \ref{ssec:friction} can be taken into account as well. If
there are no dissipative forces, $\phi'_k=0$, and the original Euler-Lagrange
equations (\ref{eq:generaleulerlagrange}) are reproduced.

The strategy to handle problems with such forces would be as follows:
First, formulate a Lagrange function $L$ with the kinetic energy and the
potential from interactions or forces that can be expressed via a potential
$U$ in convenient coordinates. Then, evaluate  the generalized forces
$\phi'_k$ for the non-conservative forces from their description in
real space via (\ref{eq:generalizedforcedefinition}). Subsequently, solve the
set of (possibly coupled) differential equations
(\ref{eq:lagrangeEOMwithfriction}).


\subsection{Constraints in the Lagrange formalism}
One of the advantages of the Lagrange formalism is that constraints in the
motion of particles can be elegantly taken care of by an adequate choice of
coordinates. In an adequate coordinate system, a constraint does not explicitly
not show up in the equation of motion, and the number of degrees of freedoms
or generalized coordinates $q_k$ is reduced by the number of constraints.

Sometimes, however, it may be better to keep more generalized
coordinates in the system of differential equations to be solved. There is a
specific method within the Euler-Lagrange formalism that can handle explicit
constraints.

Constraints between a set of coordinates are typically formulated in the form
\begin{equation}
f(\{q_k\})=0\,.
\end{equation}
An example would be the motion of a point mass on the surface of a sphere in
the usual three-dimensional space with coordinates $x,y$ and $z$. There, the
constraint to the surface of a sphere with radius $r$ is
\begin{equation}
f(x,y,x)=x^2+y^2+z^2-r^2=0\,.
\end{equation} 

In general, constraints can not only be a function of coordinates $q_k$, but
also a function of the velocities $\dot{q}_k$, or explicitly the time $t$. The 
constraints are typically classified into the following categories:

\begin{center}
\begin{tabular}{ccc}
\toprule
&scleronomic&rheonomic\\
&(fixed in time )&(time-dependent)\\ \cmidrule(r{0.5em}){2-3}
holonomic&\multirow{2}{*}{$f(\{q_k\})=0$}&\multirow{2}{*}{$f(\{q_k\},t)=0$}\\
(independent of $\dot{q}_k$)&&\\ \addlinespace[1em]
non-holonomic&\multirow{2}{*}{$f(\{q_k\},\{\dot{q}_k\})=0$}&\multirow{2}{*}{$f(\{q_k\},\{\dot{q}_k\},t)=0$}\\
(dependent on $\dot{q}_k$)&&\\ \bottomrule
\end{tabular}
\end{center}

The method of {\em undetermined Lagrange multipliers} allows taking care of
holonomic constraints, and certain non-holonomic constraints if they are of a
certain form.

To see how these can be taken care of, we recall the variational problem in
section \ref{ssec:EulerEquations}, and formulate it for two variables $y$ and
$z$ that are connected via a constraint
\begin{equation}\label{eq:constraint}
g(y,z;x)=0\,.
\end{equation} As a reminder, in the
Euler problem for the case at hand we were looking for solutions $y(x)$ and $z(x)$ which make the
functional
\begin{equation}
J=\int\limits_{x_1}^{x_2}f(y,y',z,z'; x)\,dx
\end{equation}
extremal (i.e., minimal or maximal). This was done by adding deviations $\eta_i$ with a control parameter
$\alpha$ to the desired solutions,
\begin{equation}
y(\alpha,x)=y(x)+\alpha\eta_1(x)\,,\quad z(\alpha,x)=z(x)+\alpha\eta_2(x)\,.
\end{equation}
The extremal condition for $J$ according to (\ref{eq:costfunctionmany}) for
the two functions $y$ and $z$ is then explicitly
\begin{equation}\label{eq:twoeulers}
\frac{\partial J}{\partial\alpha}=\int\limits_{x_1}^{x_2}
\left[\left(\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial
      f}{\partial y'}\right)\frac{\partial y}{\partial\alpha}+
\left(\frac{\partial f}{\partial z}-\frac{d}{dx}\frac{\partial
      f}{\partial z'}\right)\frac{\partial z}{\partial\alpha}
\right]\,dx\stackrel{!}{=}0\,,
\end{equation}
where the derivatives of $y$ and $z$ with respect to $\alpha$ are the
deviation functions $\eta_i(x)$ from the optimal path:
\begin{equation}
\eta_1(x)=\frac{\partial y}{\partial \alpha}\quad{\rm and}\quad
\eta_2(x)=\frac{\partial z}{\partial \alpha}\,.
\end{equation}
As $\partial J/\partial\alpha=0$ needed to be valid for all possible
independent deviations $\eta_i(x)$, it was necessary that both  parentheses
in (\ref{eq:twoeulers}) are vanishing identically, which led to the Euler
equations for $y$ and $z$. With constraint (\ref{eq:constraint}), however, the
deviations are not independent anymore, but connected via
\begin{equation}
\frac{dg}{d\alpha}=\frac{\partial g}{\partial y}\frac{\partial y}{\partial\alpha}+
\frac{\partial g}{\partial z}\frac{\partial z}{\partial\alpha}=
\frac{\partial g}{\partial y}\eta_1+\frac{\partial g}{\partial z}\eta_2=0\,.
\end{equation}
The deviation $\eta_2$ now can be expressed via the deviation $\eta_1$,
\begin{equation}
\eta_2=-\eta_1\frac{\partial g/\partial y}{\partial g/\partial z}\,,
\end{equation}
so the condition for an extremal $J$ turns into
\begin{equation}
\frac{\partial J}{\partial\alpha}=\int\limits_{x_1}^{x_2}
\left[\left(\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial
      f}{\partial y'}\right)-
\left(\frac{\partial f}{\partial z}-\frac{d}{dx}\frac{\partial
      f}{\partial z'}\right)\frac{\partial g/\partial y}{\partial g/\partial z}
\right]\eta_1(x)\,dx\stackrel{!}{=}0\,.
\end{equation}
For this expression to hold for all deviations $\eta_1(x)$, the square bracket
needs to vanish identically, or
\begin{equation}
\left(\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial
      f}{\partial y'}\right)\frac{1}{\partial g/\partial y}=
\left(\frac{\partial f}{\partial z}-\frac{d}{dx}\frac{\partial
      f}{\partial z'}\right)\frac{1}{\partial g/\partial z}=:-\lambda\,.
\end{equation}
As each side of this expression contains derivatives with respect to $y$ and
$y'$, or $z$ and $z'$ only, the so-called {\em Lagrange undetermined
  multiplier} $\lambda$ can at most be a function of $x$. Then,
\begin{eqnarray}\label{eq:constrainedEulerEquations}
\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}
\,+\,\lambda(x)\frac{\partial g}{\partial y}&=&0\,,\quad{\rm and}\nonumber\\
\frac{\partial f}{\partial z}-\frac{d}{dx}\frac{\partial f}{\partial z'}
\,+\,\lambda(x)\frac{\partial g}{\partial z}&=&0\,.
\end{eqnarray}
This is a set of differential equations for three functions $y(x)$, $z(x)$ and
$\lambda(x)$. Together with the constraint condition (\ref{eq:constraint}),
there are three equations for the three unknown functions.

The method can be extended to more than two functions, and more than a single
constraint. Applying this general formalism to the Hamilton principle, the
Euler-Lagrange equations of motion with $s$ constraints become
\begin{equation}\label{eq:EulerLagrangeWithConstraints}
\frac{\partial L}{\partial q_k}-\frac{d}{dt}\,\frac{\partial L}{\partial
  \dot{q}_k}\,+\,\sum\limits_{i=1}^s\lambda_i(t)\frac{\partial f_i}{\partial q_k}=0\,,
\end{equation}
with $s$ constraint equations $f_i=0$ for the dynamic variables $q_k(t)$, and
the $s$ Lagrange undetermined multipliers $\lambda_i(t)$. 

Comparing (\ref{eq:EulerLagrangeWithConstraints})  with the definition of
equations of motion under arbitrary generalized forces in
(\ref{eq:lagrangeEOMwithfriction}), one can interpret the Lagrange multipliers
directly as forces that are not captured by standard Lagrange formalism:
\begin{equation}
\frac{\partial L}{\partial q_k}-\frac{d}{dt}\,\frac{\partial L}{\partial
  \dot{q}_k}=-\sum\limits_{i=1}^s\lambda_i(t)\frac{\partial f_i}{\partial q_k}=
-\sum\limits_{i=1}^s\phi_{i,k}\quad{\rm with}\quad\phi_{i,k}:=\lambda_i(t)\frac{\partial f_i}{\partial q_k}\,.
\end{equation}
The $\phi_{i,k}$ represents the constraining force in direction of coordinate
$q_k$ due to the constraint $f_i=0$. This is useful to find out the
constraining forces in the Lagrange formalism, which tries to avoid
the calculation of constraining forces by an adequate choice of variables to
describe the problem. Such constraining forces are not interesting to
understand the dynamics of the system, but may be important
in an ``engineering'' context, where a system has to be designed to provide the
constraining forces.

\clearpage 

\section{Elements of Hamiltonian mechanics}
The introduction of the Hamilton function has a much wider use than that of a
conserved quantity in time-invariant physical systems. In a similar way as the
Lagrange function is used to derive the complete dynamics of a system, there
is a method constructing a set of equations of motion that are based on the
Hamilton 
function. This method offers some advantages in the complexity of the equation
of motion, and may lead to a better handling of cyclic variables. The
main advantage of that method, however, is in its use in other areas of
physics, namely quantum mechanics: There, the Hamilton function is at the
center of describing the dynamics of a system. Similarly, Hamiltonian
mechanics is heavily used is in statistical physics, when ensembles of
many particles, like in a fluid, are analyzed.

The starting point for this yet another approach to find equations of motion
of a 
system is the transition from a dynamic variable set $\{q_k,\dot{q}_k\}$ to a
set that is made up by the generalized coordinates, and their canonically
conjugated momenta, $\{q_k,p_k\}$. This should be possible because the momenta
$p_k$ and velocities $\dot{x}_k$ carry somewhat similar information.

We remember the definition (\ref{eq:momentumdef2}) of the generalized
momentum, 
\begin{equation}\label{eq:momentumdefinition2}
p_k=\frac{\partial L}{\partial \dot{q}_k}
\end{equation}
and also write the Lagrange equations of motion (\ref{eq:LagrangeEoM}) with
this  momentum:
\begin{equation}
\frac{\partial L}{\partial q_k}-\frac{d}{dt}\frac{\partial L}{\partial
  \dot{q}_k}=0\quad\rightarrow\quad
\dot{p}_k=\frac{\partial L}{\partial q_k}
\end{equation}
A fast way to come to equations of motion is to write the Hamilton function as
a function of the new dynamic variables (and time, if necessary):
\begin{equation}
H=H(\{q_k\}, \{p_k\}, t)\,,
\end{equation}
and try to write down the full differential $dH$ in all the variables:
\begin{equation}\label{eq:fullhamiltondifferential}
dH=\sum\limits_k\left(\frac{\partial H}{\partial q_k}\,dq_k+\frac{\partial
    H}{\partial p_k}\,dp_k\right)+\frac{\partial H}{\partial t}\,dt\,.
\end{equation}
In a next step, we try to express the total differential $dH$ from 
its definition (\ref{eq:hamiltoniandefinition}), and
identify the partial derivatives in (\ref{eq:fullhamiltondifferential}). For
this, we first simplify the Hamilton function by using definition of $p_k$,
\begin{equation}
H=\sum\limits_k\frac{\partial L}{\partial \dot{q}_k}\dot{q}_k-L=
\sum\limits_k p_k\dot{q}_k-L\,,
\end{equation}
and then calculate the total differential $dH$ from this expression:
\begin{eqnarray}\label{eq:fullhamiltondifferential2}
dH&=&d(\sum\limits_k p_k\,\dot{q}_k-L)=d\sum\limits_k p_k\,\dot{q}_k-dL\nonumber\\
&=&\sum\limits_k\left( p_k\,d\dot{q}_k + \dot{q}_k\,dp_k\right)
-\Bigg[
\sum\limits_k\underbrace{\frac{\partial L}{\partial q_k}}_{=\dot{p}_k}dq_k
+\underbrace{\frac{\partial L}{\partial \dot{q}_k}}_{=p_k}d\dot{q}_k
+\frac{\partial L}{\partial t}dt \Bigg]
\nonumber\\
&=&\sum\limits_k\left(p_k\,d\dot{q}_k+\dot{q}_k\,dp_k-\dot{p}_k\,dq_k-p_k\,d\dot{q}_k
\right)
-\frac{\partial L}{\partial t}dt\nonumber\\
&=&\sum\limits_k\left(\dot{q}_k\,dp_k-\dot{p}_k\,dq_k\right)
-\frac{\partial L}{\partial t}dt
\end{eqnarray}
In this transformation, the Lagrange equation of motion was inserted when
$\partial L/\partial q_k$ was replaced with $\dot{p}_k$. The result is a full
differential $dH$, expressed in full differentials of the parameters of $H$. By
comparing the coefficients in (\ref{eq:fullhamiltondifferential}) and
(\ref{eq:fullhamiltondifferential2}), one finds
\begin{equation}\label{eq:HamiltonEoM}
\dot{q}_k=\frac{\partial H}{\partial p_k}\,,\qquad
-\dot{p}_k=\frac{\partial H}{\partial q_k}\,,
\end{equation}
and 
\begin{equation}
-\frac{\partial L}{\partial t}=\frac{\partial H}{\partial t}\,.
\end{equation}
Equations (\ref{eq:HamiltonEoM}) are in fact already the equations of motion
for the dynamic variables $q_k$ and $p_k$, and are referred to as {\em
  Hamilton's equation of motion}.

 If there are $N$ degrees of
freedom (i. e., $N$ different indices $k$), then (\ref{eq:HamiltonEoM}) forms a
system of $2N$ ordinary differential equations of first order. They are
equivalent with the other equations of motion of a system; the Lagrange
equations of motion (\ref{eq:LagrangeEoM}) form a system of $N$ differential
equations of second order. Both sets of equations of motion require $2N$
integration constants to be solved.

Before we use the Hamiltonian approach, we look at the
total temporal derivative of the Hamilton function, which was shown to vanish
if a system was invariant to translations in time (see section
\ref{ssec:Noether3}). It can be directly calculated:
\begin{eqnarray}
\dot{H}=\frac{dH}{dt}&=&\sum\limits_k\left(
  \frac{\partial H}{\partial q_k}\dot{q}_k
  +\frac{\partial H}{\partial p_k}\dot{p}_k\right)
+\frac{\partial H}{\partial t}\nonumber\\
&=&\sum\limits_k\left(-\dot{p}_k\dot{q}_k+\dot{q}_k\dot{p}_k\right)
+\frac{\partial H}{\partial t}=\frac{\partial H}{\partial t}\,.
\end{eqnarray}
The total derivative  $\dot{H}$ is the same as the explicit time
dependency $\partial H/\partial t$, so if $H$ is not explicitly time
dependent, $H$ is a conserved quantity. As before, if $U$ is independent of
the velocities $\dot{q}_k$, and the transformations from $\{x_l\}$ to
$\{q_k\}$ is independent of time, then $H=E$.

The strategy to obtain the Hamilton equations of motion is summarized below:
\begin{itemize}
\item[1.] Express the kinetic energy $T$ in the velocities $\dot{q}$.
\item[2.] Find the potential energy $U$ as a function of the coordinates $q_k$.
\item[3.] Find the canonically conjugated momenta $p_k=\partial L/\partial
  \dot{q}_k$
\item[4a.] Express the kinetic energy $T$ in these momenta $p_k$ if this is easy,
  and obtain the Hamilton function $H=T+U$ directly if the conditions for
  $H=E$ are met (no time-dependent coordinates $q_k$, no $\dot{q}_k$ in $U$).
\item [4b.]Otherwise, obtain $H$ via the definition
  (\ref{eq:hamiltoniandefinition}).
\item [5.]Proceed to the equations of motion (\ref{eq:HamiltonEoM}).
\end{itemize}

\subsection{Simple example: harmonic oscillator}
As before, we apply the Hamiltonian formalism first to a simple
physical system, the one-dimensional harmonic oscillator. We
have seen in section \ref{ssec:harmonicosc} that
\begin{equation}
T=\frac{1}{2}m\dot{x}^2\quad{\rm and}\quad U=\frac{1}{2}kx^2\,.
\end{equation}
We use the standard Cartesian coordinate $x$ as coordinate, and find the
corresponding momentum
\begin{equation}
p=\frac{\partial L}{\partial \dot{x}}=\frac{\partial(T-U)}{\partial \dot{x}}=\frac{\partial T}{\partial \dot{x}}=m\dot{x}\,.
\end{equation}
This allows expressing the kinetic energy via the momentum,
\begin{equation}
T=\frac{p^2}{2m}\,,
\end{equation}
and since $U$ is not velocity-dependent and we work with Cartesian
coordinates, 
\begin{equation}\label{eq:harmonicH}
H=T+U=\frac{p^2}{2m}+\frac{kx^2}{2}\,.
\end{equation}
The resulting set of equations of motion obtained via
(\ref{eq:HamiltonEoM}) is then
\begin{equation}
\dot{x}=\frac{1}{m}\,p\,,\quad{\rm and}\quad\dot{p}=-kx\,.
\end{equation}
For this simple example, the two equations of motion can not be solved easier
than it used to be before, as both differential equations are coupled.

The advantages of this formalism come really out on more complex problems,
where several coordinates are cyclic. 

\subsection{Hamilton dynamics and phase space}\label{ssec:phasespace}
The transition to dynamic variables $\{q_k\}$ and $\{p_k\}$ allows
visualizing the trajectory of a system in this so-called {\em phase space}. To
do so, we first keep in mind that for many systems, the Hamilton function $H$
is a constant in time. Therefore, in a diagram that represents the function
$H(\{q_k\}, \{p_k\})$, the trajectory of the system moves along a line of
constant $H$. For example, the Hamiltonian of the harmonic oscillator had a
parabolic shape in $x$ and $p$,
\begin{equation}\label{eq:SHOhamiltonian}
H=T+U=\frac{p^2}{2m}+\frac{kx^2}{2}\,,
\end{equation}
and the contour lines for a constant energy $E$ (shown in black below) are
ellipses:
\begin{figure*}[h]
\centerline{\includegraphics{contour.eps}}
\end{figure*}

In an appropriate scaling, the time evolution of the harmonic oscillator is
just a  rotation in phase space around the origin, with a constant angular
frequency $\omega=\sqrt{k/m}$, independent of the amplitude of the
oscillation, and the trajectories in phase space are closed after a period
$T=2\pi/\omega$:
\begin{figure*}[h]
\centerline{\includegraphics{harmoscphasespace1.eps}}
\end{figure*}



Similarly to the simple picture where qualitative statements could be made on
a solution by knowing the potential energy in section
\ref{sssec:classification}, statements can be made from observing contours
with fixed total energy in phase space. First, it allows identifying the type
of solutions one can expect: If the plane of a constant energy cuts the
Hamilton function near a local minimum in phase space, the trajectory in phase
space will be closed loops, pretty much as shown in the graphics above. But
for higher energies, there may be solutions that are not bound - the plane
pendulum (see section \ref{sec:pendulum}) with coordinate $\theta$ would be
such an example, where there are oscillatory solutions as well as solutions
that increase $\theta$ monotonously in time when the pendulum rotates:
 
\centerline{\includegraphics{phasespacetrajectories.eps}}

For a given total energy $E$, some statements about maximal and minimal angles
$\theta$ can be made, or for maximal and minimal values for the corresponding
generalized momentum $p_\theta$ (which happens to be the angular momentum).

\subsection{Hamiltonian mechanics for an ensemble of systems}
The concept of phase space becomes very useful if a physical system is obeyed
in the same way by a large number of particles, like molecules in a gas. Then,
the different systems can be represented by a point moving in the same phase
space at once, and one can ask the question of how an ensemble of system
distributes over time:

\centerline{\includegraphics{phasespacepoints.eps}}

The ensemble will move in a particular way, but the arrangement of points in
phase space may change its shape or orientation over time. In such a
situation, it can be useful to introduce a density $\rho$ of systems in phase
space, that describes how many systems $dN$ can be found in a infinitesimal
phase space volume $dv$:
\begin{equation}
dN=\rho dv\,,
\end{equation}
where $dv$ is a volume in a phase space for the $s$ coordinate/momentum
pairs:
\begin{equation}
dv=(dq_1\,dq_2\,\cdots dq_s)(dp_1\,dp_2\cdots dp_s)\,.
\end{equation}
For a small volume element in phase space, one can balance how many
systems enter and leave the volume in a timer interval $dt$. For this purpose,
we consider the small area $dq\,dp$  in the in the phase space for one
coordinate $q$ and momentum $p$, and evaluate the flow of systems into this
area:

\centerline{\includegraphics{phasespacevolume.eps}}

The number of systems $dN_1$ that enter the volume in the time interval $dt$
(or the rate $dN_1/dt$) from the left is
\begin{equation}\label{eq:side1flux}
\frac{dN_1}{dt}=\rho\,\frac{dq}{dt}\,dp=\rho\,\dot{q}\,dp\,.
\end{equation}
This can be understood in the following way: The first term $\rho$ measures
how many points are there per unit phase space area, the second term
$\dot{q}$ captures how fast these points move from left to right into the
volume, and the third term $dp$ captures how wide the area is in $p$
direction where systems can enter the area under consideration. The
expression $\rho\dot{q}$ is therefore a flow density per unit of $p$. In a
similar way, the rate of systems entering via the bottom boundary is 
\begin{equation}
\frac{dN_2}{dt}=\rho\,\frac{dp}{dt}\,dq=\rho\,\dot{p}\,dq\,.
\end{equation}
To evaluate how the rate at which systems leave the test volume on the right
side, one finds the flow by Taylor expansion of the flow density $\rho\dot{q}$
in $q$ up to the first order:
\begin{equation}
\frac{dN_3}{dt}=\left(\rho\,\dot{q}+\frac{\partial}{\partial q}
  \big(\rho\,\dot{q}\big)\,dq+\ldots\right)dp\,.
\end{equation}
In a similar way, the rate of systems leaving through the top border is
\begin{equation}
\frac{dN_4}{dt}=\left(\rho\,\dot{p}+\frac{\partial}{\partial p}
  \big(\rho\,\dot{p}\big)\,dp+\ldots\right)dq\,.
\end{equation}
Summing up the net flow of systems into the area $dq\,dp$, we get
\begin{eqnarray}
  \frac{dN}{dt}&=&
  \frac{dN_1}{dt}+\frac{dN_2}{dt}-\frac{dN_3}{dt}-\frac{dN_4}{dt}\nonumber\\
&=&-\left[\frac{\partial}{\partial q}(\rho\,\dot{q})
  +\frac{\partial}{\partial p}(\rho\,\dot{p})\right]dq\,dp\nonumber\\
&=&-\left[\frac{\partial\rho}{\partial q}\dot{q}
  +\rho\frac{\partial\dot{q}}{\partial q}
  +\frac{\partial\rho}{\partial p}\dot{p}
+\rho\frac{\partial\dot{p}}{\partial p}\right]dq\,dp
\end{eqnarray}
Using the Hamilton equations of motion (\ref{eq:HamiltonEoM}) for the second
and last term,
\begin{equation}\label{eq:vanishingderivatives}
\frac{\partial\dot{q}}{\partial q}=\frac{\partial}{\partial q}
\left(\frac{\partial H}{\partial p}\right)
 =\frac{\partial^2H}{\partial q \partial p}\quad{\rm and}\quad
\frac{\partial\dot{p}}{\partial p}=\frac{\partial}{\partial  p}
\left(-\frac{\partial H}{\partial q}\right)
=-\frac{\partial^2H}{\partial q \partial p}\,,
\end{equation}
one can see that they vanish, and that
\begin{equation}
\frac{dN}{dt}=-\left[\frac{\partial\rho}{\partial q}\dot{q}
 + \frac{\partial\rho}{\partial p}\dot{p}\right]dq\,dp\,.
\end{equation}
Up to here, we have considered only the flux into the area $dq\,dp$ (or the
phase space volume) of a single coordinate/momentum pair $(q,p)$ by balancing
the flux through the sides of this volume. For $s$ degrees of
freedom (or $q_k/p_k$ pairs), the phase space volume $dv$ is not a square, but
a hypercube in $2s$ dimensions, with $2s$ surfaces. We can balance the flux
into $dv$ in a similar way as for one dimension, but need to replace the width
$dp$ of side 1 in expression (\ref{eq:side1flux}) by the ``area'' $ds_{1,k}$ of
surface $(1,k)$ of the hypercube:
\begin{equation}
dq\,dp\rightarrow ds_{1,k}=(dq_1\,dq_2\cdots dq_{k-1}\,dq_{k+1}\cdots
dq_s)(dp_1\cdots dp_s)\,.
\end{equation}
With this, the rate of systems entering via surface $(1,k)$ is
\begin{equation}
dN_{1,k}=\rho\,\dot{q}_k\,ds_{1,k}\,dt\,.
\end{equation}
The difference between the opposing hypersurfaces $(1,k)$ and $(3,k)$
becomes
\begin{eqnarray}
dN_{1,k}-dN_{3,k}&=&-\frac{\partial}{\partial
  q_k}(\rho\dot{q}_k)\,dq_k\,ds_{1,k}\,dt\nonumber\\
&=&-\frac{\partial}{\partial q_k}(\rho\dot{q}_k)\,dv\,dt\nonumber\\
&=&-\left[\frac{\partial\rho}{\partial q_k}\dot{q}_k
+\rho\frac{\partial\dot{q}_k}{\partial q_k}\right]dv\,dt
\end{eqnarray}
A similar argument holds for the opposing hypersurfaces  $(2,k)$ and
$(4,k)$. When summing up the flux through all surfaces, the terms containing 
$\partial\dot{q}_k/\partial q_k$ and $\partial\dot{p}_k/\partial p_k$ vanish
again because of (\ref{eq:vanishingderivatives}). This leads to a total number
\begin{eqnarray}
dN&=&\left(\sum\limits_kdN_{1,k}+dN_{2,k}-dN_{3,k}-dN_{4,k}\right)dt\nonumber\\
&=&-\sum\limits_k\left[\frac{\partial \rho}{\partial q_k}\dot{q}_k +
\frac{\partial \rho}{\partial p_k}\dot{p}_k\right]dv\,dt\label{eq:fluxthroughwalls}
\end{eqnarray}
of systems entering the phase space volume $dv$ in time $dt$.

Since the number of systems moving in phase space can not change, the
influx $dN$ into $dv$ must cause a change in the phase space
density $\rho$ in time:
\begin{equation}\label{eq:phasespacebytime}
dN=\frac{\partial\rho}{\partial t}\,dv\,dt\,.
\end{equation}
Comparing this with (\ref{eq:fluxthroughwalls}) and omitting $dv$ and $dt$ leads to 
\begin{equation}
\frac{\partial\rho}{\partial t}=
-\sum\limits_k\left[\frac{\partial \rho}{\partial q_k}\dot{q}_k +
\frac{\partial \rho}{\partial p_k}\dot{p}_k\right]\,,
\end{equation}
and after reordering
\begin{equation}
\frac{\partial\rho}{\partial t}+
\sum\limits_k\left[\frac{\partial \rho}{\partial q_k}\dot{q}_k +
\frac{\partial \rho}{\partial p_k}\dot{p}_k\right]=
\frac{d}{dt}\rho=0\,.
\end{equation}
This result is referred as {\em Liouville's theorem}\footnote{published in
  1838 by \href{https://en.wikipedia.org/wiki/Joseph_Liouville}{Joseph Liouville}, 1809-1882}, and states that the density $\rho$ of
points in phase space under time evolution according to Hamilton's equation
stays constant. To understand what this result means, we consider an ensemble
of systems subject to the motion of a harmonic oscillator, described by the
Hamilton function $H$ in (\ref{eq:SHOhamiltonian}). The ensemble should be
initially distributed over a large range of positions $x$ with a small spread
in momentum $p$ and thus little kinetic energy. A quarter of an oscillation
period $T=2\pi/\omega$ later, the distribution has evolved in phase space:

\begin{figure*}[h]
\centerline{\includegraphics{harmoscphasespace.eps}}
\end{figure*}

While the distribution changed its configuration in phase space, the density
in the phase space volume that initially contained the distribution did not
change, and the large spread in $p$ together with a large spread in $x$
translated to a distribution with a large spread in $p$ and a small spread in
$x$. For the special case of a harmonic oscillator, even the shape of the
distribution stays constant, but for more complex systems, the Liouville
theorem $d\rho/dt$ still holds, i.e., the {\em phase space distribution is
incompressible} as long as the evolution in time can be described by a
Hamilton function, i.e., is non-dissipative.

\subsection{Evolution of functions in time and Poisson brackets}
Often, one is interested in the time evolution of a quantity that is a
function of all coordinates and/or positions of a physical system; the center
of mass position in section \ref{ssec:COMmotion}, or the total angular momentum
in section \ref{ssec:angularmomentum} were such examples. In general, such a
function can be represented as
\begin{equation}
f=f(\{q_k\},\{p_k\},t)\,.
\end{equation}
Then, the temporal derivative  of this function in time can be calculated in
the usual way by differentiating with respect to  its dependent variables, and
using the Hamilton equations of motion (\ref{eq:HamiltonEoM}):
\begin{eqnarray}
\frac{df}{dt}&=&\frac{\partial f}{\partial t}+\sum\limits_k\left(
  \frac{\partial f}{\partial q_k}\,\dot{q}_k
  +\frac{\partial f}{\partial p_k}\,\dot{p}_k \right)\nonumber\\
&=&\frac{\partial f}{\partial t}+\sum\limits_k\left(
  \frac{\partial f}{\partial q_k}\,\frac{\partial H}{\partial p_k}
  -\frac{\partial f}{\partial p_k}\,\frac{\partial H}{\partial q_k}\right)\,.
\end{eqnarray}
By defining the so-called {\em Poisson bracket} as a short notation,
\begin{equation}
\{f,g\}:=\sum\limits_k\left(
  \frac{\partial f}{\partial q_k}\,\frac{\partial g}{\partial p_k}
  -\frac{\partial f}{\partial p_k}\,\frac{\partial g}{\partial q_k}\right)\,,
\end{equation}
the temporal derivative can be written as
\begin{equation}
\frac{df}{dt}=\frac{\partial f}{\partial t}+\{f,H\}
\end{equation}
This is a notation that resembles closely the expression for the time
evolution of an operator in a system in quantum physics. Some other properties
of these Poisson brackets that are very closely related to corresponding
expressions quantum physics can also be verified:
\begin{eqnarray}
\{q_i, p_j\}&=&\sum\limits_k\Bigg(
  \frac{\partial q_i}{\partial q_k}\,\frac{\partial p_j}{\partial p_k}
  -\underbrace{\frac{\partial q_i}{\partial p_k}}_{=0}\,
  \underbrace{\frac{\partial p_j}{\partial q_k}}_{=0}
  \Bigg)\nonumber\\
&=&\sum\limits_k\delta_{ik}\delta_{jk}=\delta_{ij}\,,
\end{eqnarray}
and similarly, for Poisson brackets between coordinates:
\begin{equation}
\{q_i, q_j\}=\sum\limits_k\Bigg(
\frac{\partial q_i}{\partial q_k}\,
\underbrace{\frac{\partial q_j}{\partial p_k}}_{=0}
  -\underbrace{\frac{\partial q_i}{\partial p_k}}_{=0}\,
  \frac{\partial q_j}{\partial q_k}
  \Bigg)=0\,,
\end{equation}
and with a similar argument, $\{p_i, p_j\}=0$. In general, two functions $f$
and $g$ are said to {\em commute with each other} if 
\begin{equation}
\{f,g\}=0\,.
\end{equation}

A useful expression can be obtained by evaluating the Poisson bracket with a
momentum $p_i$: 
\begin{eqnarray}
  \{p_i, f\}&=&\sum\limits_k\Bigg(
  \underbrace{\frac{\partial p_i}{\partial q_k}}_{=0}\,
  \frac{\partial f}{\partial p_k}\,
  -\underbrace{\frac{\partial p_i}{\partial p_k}}_{=\delta_{ik}}\,
  \frac{\partial f}{\partial q_k}\Bigg)\nonumber\\
  &=&-\sum\limits_k\Bigg(\delta_{ik}\,\frac{\partial f}{\partial q_k}\Bigg)
  =-\frac{\partial f}{\partial q_i}\,.
\end{eqnarray}
Again, a symmetric expression can be obtained for a Poisson bracket with $q_i$:
\begin{equation}
\{q_i, f\}=\frac{\partial f}{\partial p_i}\,.
\end{equation}
Poisson brackets allow compactly formulating the evolution in time and the
relation between different functions of the state of
the system characterized by a set of dynamic variables $\{q_k, p_k\}$. They
are very closely related to the commutators on quantum physics; in fact, one
of the ways to make a transition from classical mechanics to quantum physics
simply postulates particular properties of the commutators between operators
that differ from classical physics, and take over the rest of the dynamics
from the tool set developed in classical Hamiltonian mechanics.

\clearpage
\section{Central force motion and two-body problem}
One of the both historically and practically important problems in classical
mechanics is the dynamics of two objects that exert a central force on each
other. As a reminder, central forces were defined as forces that two point
masses exert on each other along their distance vector:

\begin{figure*}[h]
\centerline{\includegraphics{centralforces.eps}}
\end{figure*}

Such forces are found in the gravitational interaction between celestial
bodies and in the Coulomb interaction between charged particles, and often
are in good approximation not subject to dissipation.

\subsection{Simplification through conservation laws}\label{ssec:planemotion}
Central forces only depend on the difference vector between the positions, and
thus can be derived from a potential that only depends on the distance vector:
\begin{equation}
\vec{F}=\vec{F}(\vec{r}_1-\vec{r}_2)\quad{\rm and}\quad
U=U(\vec{r}_1-\vec{r}_2)\,.
\end{equation}
Furthermore, since the form of the potential does not depend on the
orientation of the distance vector $\vec{r}=\vec{r}_1-\vec{r}_2$ between the two
masses, the potential is only a function of the distance $r=|\vec{r}|$, so
$U=U(|\vec{r}|)=U(r)$. The Lagrange function of the system is given by
\begin{equation}
L=T-U=\frac{1}{2}m_1\dot\vec{r}_1^2+\frac{1}{2}m_2\dot\vec{r}_2^2 
 - U(r)\,.
\end{equation}
As we have seen in section \ref{ssec:COMmotion}, a system of particles with
central forces between the particles without external forces has a constant
total momentum, and the motion of the center-of-mass position
\begin{equation}
\vec{R}=\frac{1}{m_1+m_2}(m_1\vec{r}_1+m_2\vec{r}_2)
\end{equation}
is uniform, i.e., $\ddot\vec{R}=0$. Therefore, we can choose an inertial
system where $\dot\vec{R}=0$. Next, we place the coordinate origin into the
center-of-mass position, so $\vec{R}=0$. 
We then we express the positions $\vec{r}_1, \vec{r}_2$ via the distance vector $\vec{r}$:
\begin{equation}
\vec{r}_1=\frac{m_2}{m_1+m_2}\,\vec{r}\quad{\rm and}\quad
\vec{r}_2=-\frac{m_1}{m_1+m_2}\,\vec{r}\,.
\end{equation}
With this, the Lagrange function of the problem becomes
\begin{equation}\label{eq:reducedmassdefinition}
L=\frac{1}{2}\mu\dot\vec{r}^2 - U(|\vec{r}|)\quad{\rm with}\quad
\mu:=\frac{m_1m_2}{m_1+m_2}\,.
\end{equation}
The quantity $\mu$ is referred to as the {\em reduced mass} of the two-body
problem. With this, the original two-body problem has been reduced to a single
body problem, with only the distance vector $\vec{r}$ as a dynamic
variable.

The Lagrange function does not depend on the orientation of $\vec{r}$, only
its modulus $|\vec{r}|$ via $U(|\vec{r}|)=U(r)$. This means that the problem
is spherically symmetric, so the Noether theorem in
\ref{ssec:noether-rotation} implies that the angular momentum
\begin{equation}
\vec{L}=\vec{r}\times\vec{p}
\end{equation}
is constant in time. By definition, vectors $\vec{r}$ and $\vec{p}$ are
perpendicular to  $\vec{L}$, so they stay in a fixed plane
perpendicular to the constant $\vec{L}$.
% TODO: ADD PICTURE WITH L, r, p!!
This reduces the problem  to a two-dimensional problem in this plane. An adequate set of
coordinates are polar coordinates $(r,\theta)$ with
\begin{equation}
\vec{r}=r\left(\begin{array}{c}\cos\theta\\\sin\theta\end{array}\right)
\quad{\rm and}\quad
\dot\vec{r}=
\dot{r}\left(\begin{array}{c}\cos\theta\\\sin\theta\end{array}\right)
+r\left(\begin{array}{c}-\sin\theta\\\cos\theta\end{array}\right)\dot{\theta}
\end{equation}
in the plane perpendicular to $\vec{L}$. With this, $\dot{\vec{r}}^2$ in
(\ref{eq:reducedmassdefinition}) becomes
\begin{equation}
\dot{\vec{r}}^2=\dot{r}^2+r^2\dot\theta^2\,,
\end{equation}
and the Lagrange function simplifies to
\begin{equation}
L=\frac{1}{2}\mu(\dot{r}^2+r^2\dot\theta^2)-U(r)=L(r,\dot{r},\dot{\theta})\,.
\end{equation}
As $\theta$ is a cyclic coordinate, the corresponding Lagrange equation
(\ref{eq:generaleulerlagrange}) for
$\theta$ states just that the corresponding angular momentum
\begin{equation}
p_\theta=\frac{\partial L}{\partial\theta}=\mu r^2\dot\theta=const. =:l
\end{equation}
is a constant of motion or a {\em first integral}.
% As $p_\theta$ is the
%corresponding momentum to an angle variable in the plane of motion, it is the
%constant angular momentum component the direction out of the plane of motion.
This constant angular momentum $l=|\vec{L}|$ has a simple geometric interpretation. The area
$dA$ that the radius vector $\vec{r}$ covers in a time interval $dt$ can be
calculated via
\begin{equation}
\raisebox{-6mm}{\includegraphics{arealaw.eps}}
\quad dA=\frac{1}{2}r(rd\theta)=\frac{1}{2}r^2d\theta\,,
\end{equation}
so the change of this are in time is given by
\begin{equation}\label{eq:keplerssecondlaw}
\frac{dA}{dt}=\frac{1}{2}r^2\dot\theta=\frac{l}{2\mu}=const.
\end{equation}
This is referred to as {\em
  Kepler's second law\/}\footnote{published in 1609 by
  \href{https://en.wikipedia.org/wiki/Johannes_Kepler}{Johannes Kepler},
  1571-1630}, which at that time was a heuristic description of the positions
of the planet Mars recorded by 
T.~Brahe\footnote{\href{https://en.wikipedia.org/wiki/Tycho_Brahe}{Tycho
    Brahe}, 1546-1601}. While first found with the gravitational interaction,
this law holds for any form of the potential 
$U(r)$ in a central force problem, as it is only a consequence of angular
momentum conservation.

The remaining equation of motion is for the distance $r$ between the two
bodies,
\begin{eqnarray}
\frac{\partial L}{\partial r}-\frac{d}{dt}\frac{\partial L}{\partial\dot{r}}
&=&\mu r\dot{\theta}^2-\frac{\partial U}{\partial r}-\frac{d}{dt}(\mu
\dot{r})\nonumber\\
&=&\frac{l^2}{\mu r^3}-\frac{\partial U}{\partial r}-\mu\ddot{r}=0
\end{eqnarray}
or
\begin{equation}\label{eq:radialEoM}
\ddot{r}-\frac{l^2}{\mu^2 r^3}+\frac{1}{\mu}\frac{\partial U}{\partial r}=0\,.
\end{equation}
This is an ordinary differential equation in $r$ that does not have a simple
analytical solution for a general potential $U(r)$. However, a number of
qualitative observations can be made by looking at the total energy $E$ of the
system (which is conserved, as $L$ is independent of $t$):
\begin{equation}\label{keplertotalenergy}
E=T+U=\frac{1}{2}\mu \dot{r}^2+\frac{l^2}{2\mu r^2}+U(r)\,.
\end{equation}
This expression can be resolved for the radial velocity $\dot{r}$,
\begin{equation}\label{eq:radialvelocity}
\dot{r}=\frac{dr}{dt}
=\pm\sqrt{\frac{2}{\mu}\left(E-U(r)-\frac{l^2}{2\mu r^2}\right)}\,.
\end{equation}
Collecting all terms that depend on $r$ on one side and subsequent direct
integration leads to
\begin{equation}\label{eq:centralprobperiode}
t=\int\limits_{r_0}^{r}\frac{dr'}{\sqrt{\frac{2}{\mu}\left(E-U(r')-\frac{l^2}{2\mu
      r'^2}\right)}}=t(r,r_0)\,,
\end{equation}
where $r_0$ is the distance at $t=0$. If the integral can be carried out
explicitly, The result can be inverted for the solution $r(t,r_0)$.

\subsection{Effective potential in central force motion}\label{ssec:effectivepotential}
Before moving on to a solution of the radial motion, it is helpful to group
the expression (\ref{keplertotalenergy}) into a part dependent on the radial
velocity $\dot{r}$, and the rest:
\begin{equation}
E=\frac{1}{2}\mu \dot{r}^2\,\,+\,\,\underbrace{\frac{l^2}{2\mu
    r^2}+U(r)}_{=:V_{\rm eff}(r)}\,.
\end{equation}
The so-called {\em effective Potential} $V_{\rm eff}(r)$ captures both the
original potential $U(r)$, and the kinetic energy $l^2/2\mu r^2$ associated
with the angular momentum (sometimes
referred to as {\em  centrifugal energy}), which diverges to positive values for
$r\rightarrow0$.  In a similar way that comparing of the potential with the
total energy in (\ref{sssec:classification}) allowed a classification of
solutions, one can use the effective potential to make statements on the
radial motion of the system. We consider the case of an attractive potential
$U(r)=-k/r$ with some $l>0$:

\begin{figure*}[h]
\centerline{\includegraphics{effectivepotential.eps}}
\end{figure*}

For $E=E_1$, the system is in a state where all the energy is taken up by the
effective potential energy, and nothing is left for any
radial motion. Therefore, $\dot{r}=0$, so $r$ is constant in time. This does
not mean that there is no  kinetic energy -- there is still the motion
connected with the  angular momentum $l$ as part of the
effective potential $V_{\rm eff}$. The fixed $r$ implies a circular trajectory
(with radius $r_1$), i.e.,  the system is in a bound state.

For $E=E_2$, the system is also in a bound state, but
there is enough energy to allow for radial motion, oscillating between two
extremal radii $r_{2 \rm min}$ and  $r_{2 \rm max}$
with an oscillation period that can be calculated via
(\ref{eq:centralprobperiode}). The trajectory or {\em orbit} $\vec{r}(t)$  could look like this:\\
%\begin{figure*}[h]
\centerline{\includegraphics[width=6.3cm]{centralmotion1.eps}}
%\end{figure*}

The motion in radial direction and the one due to the angular momentum $l$
are not necessarily synchronized for all potentials $U(r)$ -- in the figure
above, they are not. In this case the orbit is referred to as {\em open}.

For $E=E_3$, there is only one intersection of $V_{\rm eff}(r)$ with the line
of constant energy 
at $r_{3 \rm min}$, which means that the particle moves from infinity to the
point of closest proximity to the origin (corresponding to the center of mass
of the system), and then leaves again with
$r(t\rightarrow+\infty)\rightarrow\infty$. Because
$U(r\rightarrow\infty)\rightarrow0$, the particle approaches the center from
an asymptotically uniform
motion, interacts with the other particle and then escapes,
approaching a uniform motion into a different direction:

\begin{figure*}[h]
\centerline{\includegraphics{scatteringtrace.eps}}
\end{figure*}

\noindent This situation is referred to as a {\em scattering
  process}, and is an important concept in atomic, molecular and particle
physics.

We now come back to the problem of solving the radial motion in
(\ref{eq:radialEoM}). To assess the shape of orbits without considering the explicit time dependency, we convert (\ref{eq:radialEoM}) into an equation for
$r(\theta)$. For this, we introduce the substitution 
\begin{equation}u:=\frac{1}{r}\end{equation}
and calculate its derivatives with respect to $\theta$ (using $l=\mu r^2\dot\theta$):
\begin{equation}
\frac{du}{d\theta}=-\frac{1}{r^2}\frac{dr}{d\theta}
=-\frac{1}{r^2}\frac{dr}{dt}\frac{dt}{d\theta}
=-\frac{1}{r^2}\,\dot{r}\frac{1}{\dot\theta}=-\dot{r}\,\frac{\mu}{l}\,,
\end{equation}
and
\begin{equation}
\frac{d^2u}{d\theta^2}=-\frac{\mu}{l}\,\frac{d\dot{r}}{d\theta}
=-\frac{\mu}{l}\,\frac{d\dot{r}}{dt}\frac{dt}{d\theta}
=-\frac{\mu}{l}\,\frac{1}{\dot\theta}\,\ddot{r}
=-\frac{\mu^2}{l^2}\,r^2\,\ddot{r}\,.
\end{equation}
This can be used to substitute $\ddot{r}$ in (\ref{eq:radialEoM}) by
\begin{equation}
\ddot{r}=\frac{d^2u}{d\theta^2}\,\left(-\frac{l^2}{\mu^2r^2}\right)\,,
\end{equation}
so the differential equation (\ref{eq:radialEoM}) becomes
\begin{eqnarray}\label{eq:radialEQ2}
-\frac{l^2}{\mu r^2}\,\frac{d^2u}{d\theta^2}-\frac{l^2}{\mu r^4}\,r
&=&-\frac{\partial U}{\partial r}\quad {\rm or}\nonumber\\
\frac{d^2u}{d\theta^2}+u&=&\frac{\mu}{l^2}\,r^2\frac{\partial U}{\partial r}
\end{eqnarray}
This is a simpler differential equation that can be solved for the important
class of potentials where $U\propto1/r$.
\subsection{Kepler problems}\label{ssec:keplerproblems}
Two-body problems with a potential
\begin{equation}\label{eq:keplerpotential}
U=-\frac{k}{r}
\end{equation}
are referred to as {\em Kepler problems}, because they are equivalent to
solving the planetary motion problem. For gravitational attraction,
\begin{equation}k=Gm_1m_2\,,\end{equation}
with the gravitational constant $G$ and the heavy masses $m_1$ and $m_2$. For
the coulomb interaction,
\begin{equation}k=-\frac{q_1q_2}{4\pi\epsilon_0}\,,\end{equation}
for two charges $q_1, q_2$, with the electrical permittivity $\epsilon_0$. The
minus sign takes into account that charges of the same sign repel each other,
while charges of opposite sign attract each other. The potential
(\ref{eq:keplerpotential}) leads 
to a radial derivative $\partial U/\partial r\propto r^{-2}$, which simplifies
the differential equation (\ref{eq:radialEQ2}) to
\begin{equation}\label{eq:simpleKeplereq}
\frac{d^2u}{d\theta^2}+u-c=0\quad{\rm with}\quad
c=\frac{\mu}{l^2}\,k=const.
\end{equation}
This is a simple linear differential equation with special solutions of the form
\begin{equation}
u(\theta)=Ae^{B\theta}+c\,,
\end{equation}
with the same $c$ as in (\ref{eq:simpleKeplereq}).
Inserting this solution into (\ref{eq:simpleKeplereq}) leads to $B^2=-1$, or a
general solution
\begin{equation}\label{eq:keplersolutionsgeneral}
u(\theta)=A'e^{i\theta}+A''e^{-i\theta}+c\,,\qquad{\rm or}\quad
u(\theta)=A'''\cos(\theta-\theta_0)+c\,.
\end{equation}
with free integration constants $A'$ and $A''$, or $A'''$ and $\theta_0$
that are suitable for a description of the real-valued solution. We can choose
$\theta_0=0$, because this only orients our polar coordinate system in a
particular way. The integration constant $A'''$ needs then to be determined
form the 
initial conditions. Re-arranging the solution into a different form yields
\begin{equation}
\frac{u}{c}=\frac{1}{c}\,\frac{1}{r}=\frac{A'''}{c}\cos\theta+1\,.
\end{equation}
With a transition to the the common definitions
\begin{equation}\label{eq:fixalpha}
\alpha:=\frac{1}{c}=\frac{l^2}{\mu k}\quad{\rm and}\quad
\epsilon:=\frac{A'''}{c}\,,
\end{equation}
the solution takes the form
\begin{equation}\label{eq:conetrajectory}
\frac{\alpha}{r}=1+\epsilon\cos\theta\,.
\end{equation}
This is the parametric description of a number of curves referred to as {\em
  cone intersections}, and the dimensionless constant $\epsilon$ in this
expression is referred to as the {\em eccentricity} of the curve.
\subsubsection{Classification of orbits}
The cone intersections as solutions for the Kepler problem can be classified
according to the values of the eccentricity $\epsilon$; in the graph below,
$\alpha$ was chosen such that all trajectories have the same
distance of the nearest point $P$ on the trajectory to the origin $F$:

\begin{figure*}[h]
\centerline{\includegraphics{coneintersections.eps}}
\end{figure*}

\begin{itemize}
\item For $\epsilon=0$, the solution for $r(\theta)=\alpha$ is independent of
$\theta$, and the trajectory is a circle around the coordinate origin. This
corresponds to the bound state with the lowest total energy for a given
angular momentum $l$.
\item For  $0<\epsilon<1$, the trajectory $r(\theta)$ forms an ellipse, with
  the coordinate origin in one of its focal points $F$. The trajectory
  represents again a bound state. Point $P$ on the trajectory is
the one with the closest distance to the coordinate origin, and therfore the
shortest distance between the two bodies. It is referred to as {\em
  pericenter}, or, if reference is made to a planet orbiting the Sun, as {\em
  perihelion}, or as {\em perigee} if reference is made to a satellite
orbiting the earth. Similarly, the position $A$ on the trajectory which is
furthest away from the coordinate origin is referred to as {\em apocenter}, {\em
  apohelion}, or {\em apogee}, respectively.
If one of the two masses is much heavier than the other one, its distance from
the center of mass of the system (i.e the coordinate origin) is very
small, and therefore would be located near the focal point $F$ of the
elliptical orbit. This is {\em Kepler's first law} for planetary motion,
stating that planetary orbits are ellipses, with the Sun in one of its foci.

\item For $\epsilon=1$, the solution $r(\theta)$ represents a parabola. This
corresponds to the situation where $E=0$. It is
not a bound state anymore, because
$r(t\rightarrow\pm\infty)\rightarrow\infty$.
\item For $\epsilon>1$, the trajectories are also not bounded. Furthermore,
  the range of $\theta$ is limited because expression  
(\ref{eq:conetrajectory}) diverges for $\theta\rightarrow\theta_{\rm
  max}$, with 
\begin{equation}\label{eq:maxangle}
\cos\theta_{\rm max}=-\frac{1}{\epsilon}\,.
\end{equation}
In this case, the trajectory starts out with $\theta=\theta_{\rm max}$ and
$r\rightarrow\infty$, moves towards the coordinate center, with a decreasing
$\theta$, reaches the  pericenter $P$ for $\theta=0$, and leaves for
$r\rightarrow\infty$ with $\theta\rightarrow-\theta_{\rm max}$ (or the other
way round). Such a problem  is referred to as a {\em scattering problem}.
\end{itemize}

So far, we have discussed the solutions (\ref{eq:conetrajectory}) to the
Kepler problem only qualitatively. We still need to make the connection of the
orbit parameter $\epsilon$ with the physical properties of the two
body problem, since $\alpha$ is already fixed via (\ref{eq:fixalpha}). 
%The eccentricity $\epsilon$ should be fixed by specifying the angular momentum
%$l$ and the total energy $E$. 

The connection is easily made by using expression (\ref{keplertotalenergy})
for the total energy, and inserting the Kepler potential:
\begin{equation}
E=T+U=\frac{1}{2}\mu\dot{r}^2+\frac{1}{2}\frac{l^2}{\mu r^2}-\frac{k}{r}
\end{equation}
By definition, the radial velocity $\dot{r}$ a the pericenter $r=r_{\rm min}$
vanishes
%, and all the energy is provided by the potential and the centrifugal energy:
\begin{equation}\label{eq:totalenergykepler}
  E=\frac{1}{2}\frac{l^2}{\mu r_{\rm min}^2}-\frac{k}{r_{\rm min}}
\end{equation}
On the other hand, we have an expression for $r_{\rm min}$ from the  Kepler
solution (\ref{eq:conetrajectory}):
\begin{equation}
  r_{\rm min}=\frac{\alpha}{1+\epsilon}\,.
\end{equation}
 Inserting this into the expression for the total energy
 (\ref{eq:totalenergykepler}) yields
 \begin{eqnarray}
 E&=&\frac{1}{2}\frac{l^2}{\mu}\frac{(1+\epsilon)^2}{\alpha^2}-k\frac{1+\epsilon}{\alpha}\nonumber\\
 &=&\frac{k}{\alpha}\left[\frac{1}{2}(1+\epsilon)^2-(1+\epsilon)\right]
=\frac{k}{2\alpha}\left[\epsilon^2-1\right]\,.
\end{eqnarray}
This expression can be inverted into
\begin{equation}\label{eq:eccentricity}
\epsilon=\sqrt{\frac{2\alpha E}{k}+1}=\sqrt{1+E\frac{2l^2}{\mu k^2}}\,.
\end{equation}
With this, the orbit parameters $\alpha, \epsilon$ are fully determined by the
orbital momentum 
$l$ and the total energy $E$ of the two-body problem. Alternatively, a
particular velocity at a position in space could have been specified as an
initial condition, but $l$ and $E$ can always easily be calculated from them.

As a last step in this section, we want to apply Kepler's second law (\ref{eq:keplerssecondlaw}), and
evaluate the total orbital period $\tau$ of a bound state by equating the area covered by the radius vector
per unit of time with the ratio of the ellipse area $A$ and $\tau$:
\begin{equation}
\frac{dA}{dt}=\frac{l}{2\mu}\stackrel{!}{=}\frac{A}{\tau}=\frac{\pi a b}{\tau}\,,
\end{equation}
where $a$ and $b$ are the two semi-axes of an elliptical or circular orbit.

\begin{figure*}[h]
\centerline{\includegraphics{ellipse.eps}}
\end{figure*}

 One can easily show that $a$ and $b$ are connected with the parameters
$\epsilon$ and $\alpha$ of the cone intersection  expression
(\ref{eq:conetrajectory}):
\begin{equation}
a=\frac{\alpha}{1-\epsilon^2}\quad{\rm and}\quad b=\sqrt{\alpha a}\,.
\end{equation}
With this, one finds with $l=\sqrt{\alpha\mu k}$ from (\ref{eq:fixalpha})
\begin{equation}
\tau=\pi a b\frac{2\mu}{l}=\pi a^{3/2}\sqrt{\alpha}\frac{2\mu}{\sqrt{\alpha\mu
    k}}
=a^{3/2}\pi\sqrt{\frac{4\mu}{k}}\,.
\end{equation}
This leads to 
\begin{equation}
\frac{\tau^2}{a^3}=\frac{4\pi^2\mu}{k}=const.\,,
\end{equation}
which to a good approximation is {\em Kepler's third law}, stating that the
ratio of $\tau^2$ and the cube of the semi-major axis $a$ of a planetary orbit
is a constant for all planets orbiting around the sun. The latter can be seen
by recalling the expressions for the reduced mass $\mu$ from
(\ref{eq:reducedmassdefinition}) and $k=Gm_1m_2$ for a gravitational
potential:
\begin{equation}
\frac{4\pi^2\mu}{k}=4\pi^2\frac{m_1m_2}{m_1+m_2}\,\frac{1}{Gm_1m_2}=\frac{4\pi^2}{G(m_1+m_2)}\approx\frac{4\pi^2}{Gm_2}\,,
\end{equation}
since the mass $m_2$ of the Sun is much larger the mass $m_1$ of any planet.

\clearpage
\subsubsection{Orientation of elliptical orbits: Laplace-Runge-Lentz vector}
In the previous section, we have seen some aspects of stable orbits for Kepler
problems with
\begin{equation}
U(\vec{r})=-\frac{k}{r}\quad{\rm and}\quad\vec{F}=-\nabla
U=-\frac{k}{r^3}\vec{r}\,.
\end{equation}
One of the consequences of the conservation of angular momentum $\vec{L}$ was
that the motion takes place in a plane only. Implicit in the solutions for the
elliptical orbits was that the semi-major axis is a constant in time, i.e., its
orientation in space is fixed. When determining the integration constants in
(\ref{eq:keplersolutionsgeneral}), we chose $\theta_0=0$, and aligned the
coordinate system with the semi-major axis.

In practice, the orientation of the semi-major axis (and even the whole
equations of motion) can be obtained from a specific vector. We define
\begin{equation}\label{eq:LRLvectordef}
\vec{A}:=\vec{p}\times\vec{L}-\mu k \frac{1}{r}\vec{r}\,,
\end{equation}
with the momentum $\vec{p}$, angular momentum $\vec{L}=\vec{r}\times\vec{p}$
and position vector $\vec{r}$. This quantity is referred to as {\em
  \href{https://en.wikipedia.org/wiki/Laplace-Runge-Lenz_vector}{Laplace-Runge-Lentz  vector}}, or sometimes only as Runge-Lentz vector. It
lies in the plane of motion (because the first component of
(\ref{eq:LRLvectordef}) is perpendicular to
$\vec{L}$, and the second component is in the plane of motion anyways).
To show that this vector is a constant of motion, we calculate at the temporal
derivative of its first component:
\begin{equation}
\frac{d}{dt}(\vec{p}\times\vec{L})=
\dot\vec{p}\times\vec{L}+\vec{p}\times\underbrace{\dot\vec{L}}_{=0}
\end{equation}
Using $\dot\vec{p}=\vec{F}=-k\mu\,\vec{r}/r^3$ and $\vec{p}=\mu\,\dot\vec{r}$, we can continue
\begin{eqnarray}
\frac{d}{dt}(\vec{p}\times\vec{L})&=&
-\frac{k\mu}{r^3}(\vec{r}\times(\vec{r}\times\dot\vec{r}))\nonumber\\
&=&-\frac{k\mu}{r^3}((\vec{r}\cdot\dot\vec{r})\vec{r}-(\vec{r}\cdot\vec{r})\dot\vec{r})\nonumber\\
&=&-\frac{k\mu}{r^3}(r\dot{r}\,\vec{r}-r^2\,\dot\vec{r})\nonumber\\
&=&k\mu\left(-\frac{1}{r^2}\dot{r}\,\vec{r}+\frac{1}{r}\dot\vec{r}\right)=
k\mu\frac{d}{dt}\left(\frac{1}{r}\vec{r}\right)\,.
\end{eqnarray}
The step from the second to thrid line can be seen by writing
$\vec{r}=r\,\vec{e}_r$, differentiating this product with respect to time to get
$\dot\vec{e}_r$, and using $\vec{e}_r\perp\dot{\vec{e}}_r$. Then, the
expression can be re-arranged into 
\begin{equation}
\frac{d}{dt}\left(\vec{p}\times\vec{L}-k\mu\frac{1}{r}\vec{r}\right)=
\frac{d}{dt}\vec{A}=0\,.
\end{equation}
Thus, the Laplace-Runge-Lentz vector is conserved over time. Its orientation with
respect to the elliptical motion can be seen by forming the scalar product
with the position vector $\vec{r}$:
\begin{eqnarray}
\vec{A}\cdot\vec{r}&=&\vec{r}\cdot(\vec{p}\times\vec{L})-\mu k
\frac{\vec{r}\cdot\vec{r}}{r}\nonumber\\
&=&\vec{L}\cdot\underbrace{(\vec{r}\times\vec{p})}_{=\vec{L}}-\mu k r
=l^2-\mu k r\,.
\end{eqnarray}
On the other side, $\vec{A}\cdot\vec{r}=A \,r \cos\theta$, where $A=|\vec{A}|$
and $\theta$ is the angle between vectors $\vec{A}$ and $\vec{r}$. Reordering
this expression into
\begin{equation}\label{eq:rungelentzscalar}
\frac{l^2}{\mu k r}=1+\frac{A}{\mu k}\cos{\theta}
\end{equation}
suggests that the angle $\theta$ here is the same as in the solutions for the
cone intersections in (\ref{eq:conetrajectory}). Thus, for $\theta=0$, the
vector $\vec{A}$ points in the same direction as $\vec{r}$, namely in the
direction of the pericenter, parallel to the semi-major axis of the
ellipse:

\begin{figure*}[h]
\centerline{\includegraphics{rungelentzvector.eps}}
\end{figure*}

Direct comparison of (\ref{eq:rungelentzscalar}) with the expression for the
cone intersections (\ref{eq:conetrajectory}) also reveals the connection
between the length of the Laplace-Runge-Lentz vector and the eccentricity:
\begin{equation}
A=\mu k\epsilon
\end{equation}

\clearpage
\subsection{Scattering orbits}
We now come back to the orbits in section (\ref{ssec:effectivepotential})
corresponding to a total energy $E>0$, such that $\vec{r}(t)$
is not bounded, i.e., $r(t\rightarrow\pm\infty)\rightarrow\infty$. This is
also the only type of solutions for potentials modeling the two-body
interaction that are repulsive:

%\begin{figure*}[h]
\centerline{\includegraphics{effectivepotential2.eps}}
%\end{figure*}

Typically, the interactions $U(r)$ vanish for large $r$; there, the particle
is in uniform motion with a constant velocity $\dot\vec{r}=\vec{v}'_0$. The
prime with the velocity should indicate that reference is made to a relative
velocity between the two bodies, because we still are in the framework of an
effective one-body problem. The vector $\vec{r}$ will move towards the
pericenter $P$ with a minimal distance from the center-of-mass position (or
coordinate origin $O$), and then
move away again. For $r\rightarrow\infty$, the motion becomes uniform again,
with a new constant velocity $\vec{v}'_1$. As we are still looking at a
conservative interaction, the velocities have the same modulus, but a
different direction. Asymptotically, the effect of the two-body interaction is
a deflection of the particle by an angle $\varphi'$. The geometry of that
interaction is shown in the diagram below:

\begin{figure*}[h]
\centerline{\includegraphics{scattering1.eps}}
\end{figure*}

First, we note that the {\em scattering angle} $\varphi'$ between the asymptotic
velocities $\vec{v}'_0$ and $\vec{v}'_1$ is determined by the maximal angle
$\theta_{\rm max}$ in the polar coordinate system (oriented in the direction
$O-P$ in the figure):
\begin{equation}\label{eq:deflectionangle}
\varphi'=\pi-2\theta_{\rm max}
\end{equation} 

For any two-body central force problem, the angular momentum is a conserved
quantity. We therefore try to evaluate it from the geometric parameters shown
in the above diagram, using 
\begin{equation}
l=|\vec{L}|=|\vec{r}\times\vec{p}|=\mu|\vec{r}\times\vec{v}'_0|=\mu\,r\,
v_0'\,\sin\alpha=\mu\,r\,v_0'\,\sin(\pi-\alpha)\,,
\end{equation}
where here, $\alpha$ is the angle between $\vec{r}$ and
$\vec{v}_0'$. Asymptotically, the expression 
\begin{equation}
r\sin(\pi-\alpha)=:b
\end{equation}
measures the shortest distance $b$ between the trajectory the particle would
have taken without interaction, and the coordinate origin. The distance $b$ is
referred to as {\em impact parameter} of the scattering trajectory, and
measures, casually speaking, how far the scattering center $O$ was
missed if there were no interaction. With this, one can fix the angular
momentum to 
\begin{equation}\label{eq:lfromimpact}
l=\mu\,v_0'\,b
\end{equation}
The diagram above shows a scattering problem for a repulsive potential. By
convention, the scattering angle $\varphi'$ in this case is counted
positively. For a scattering problem with an attractive potential the
scattering angle is negative, compatible with definition (\ref{eq:deflectionangle}):

\begin{figure*}[h]
\centerline{\includegraphics{scattering2.eps}}
\end{figure*}

Since the scattering angle $\varphi'$ is a simple function
(\ref{eq:deflectionangle}) of the maximum angle
$\theta_{\rm max}$ of the trajectory in a polar coordinate system with the
$\theta=0$ direction defined by the direction $O-P$, we try to evaluate this
angle for a general potential $U(r)$. For this, we go back to the expression
(\ref{eq:radialvelocity}) for the radial velocity in a central force problem:
\begin{equation}
\dot{r}=\frac{dr}{dt}
=\pm\sqrt{\frac{2}{\mu}\left(E-U(r)-\frac{l^2}{2\mu r^2}\right)}\,.
\end{equation}
We are interested in an expression $\theta(r)$, so we use the above expression
to construct a useful derivative:
\begin{eqnarray}
\frac{d\theta}{d r}&=&\frac{d\theta}{d t}\,\frac{d t}{d
  r}=\frac{\dot\theta}{\dot{r}}\nonumber\\
&=&\frac{l}{\mu r^2}\,\frac{1}{\dot{r}}=\frac{l}{\mu r^2}\,\frac{1}{
\pm\sqrt{\cdots}}\,.
\end{eqnarray}
Sorting again components that depend on $r$ on one side leads to
\begin{equation}
d\theta=\frac{l}{\mu r^2}\,\frac{dr}{\pm\sqrt{\cdots}}\,,
\end{equation}
which we integrate from the pericenter (at $\theta=0$) to infinity. Since the
distance $r$ increases monotonously, we can choose the positive sign (assuming
$l>0$), and get an expression for $\theta_{\rm max}$:
\begin{equation}\label{eq:maximumtheta}
\theta_{\rm max}=\int\limits_{r_{\rm min}}^{+\infty}\frac{l\,dr}{
\mu r^2\sqrt{\frac{2}{\mu}\left(E-U(r)-\frac{l^2}{2\mu r^2}\right)}}\,.
\end{equation}
To carry out this integration, we still need to know the $r_{\rm min}$, which
can be obtained from (\ref{eq:radialvelocity}), knowing that at $r_{\rm
  min}$, the radial velocity $\dot{r}$ vanishes:
\begin{equation}
E-U(r_{\rm min})-\frac{l}{2\mu\, r_{\rm min}^2}=0\,.
\end{equation}
This equation needs to be solved for the integration boundary $r_{\rm min}$
in  (\ref{eq:maximumtheta}), leading to the maximal polar angle $\theta_{\rm
  max}= \theta_{\rm 
  max}(E,l)$. The energy $E$ in the system is the energy in
the center-of-mass system, so we should note it as 
\begin{equation}
E'=\frac{\mu v_0'^2}{2}\,.
\end{equation}
As the angular momentum in a scattering problem is conveniently given by the
impact parameter $b$ via (\ref{eq:lfromimpact}), the scattering angle
$\varphi'$ becomes only a function of $b$ and $v'_0$, or $b$ and $E'$.

\subsubsection{Scattering  angle for Kepler problems}
For general potentials $U(r)$, it can be difficult or impossible to obtain
the maximal polar angle $\theta_{\rm max}$ from (\ref{eq:maximumtheta}) in a
closed form. For the important Kepler problems introduced in section
\ref{ssec:keplerproblems} it is
possible, but it is even easier to use the result (\ref{eq:maxangle}):
\begin{equation}\label{eq:maxangle2}
\cos\theta_{\rm max}=-\frac{1}{\epsilon}\,,
\end{equation}
with the eccentricity $\epsilon$ given by (\ref{eq:eccentricity}). In the
parametric equation (\ref{eq:conetrajectory}) for the Kepler orbits, we
implicitly used $k>0$ corresponding to attractive 
interactions to obtain a radial distance $r>0$. However, the sign of the
integration constant $\alpha$ in 
(\ref{eq:fixalpha}) can be changed for $k<0$ without affecting any of the
derivations, resulting in $r>0$ for repulsive
interactions, e.g. of two electric charges of the same sign. Expression
(\ref{eq:eccentricity}) for $ \epsilon$ is not
affected by this sign change, so $\theta_{\rm max}$  can be obtained via
(\ref{eq:maxangle2}), resulting in an expression for the scattering angle $\varphi'$:
\begin{equation}
-\frac{1}{\epsilon}=\cos\theta_{\rm max}=\cos\left(\frac{\pi}{2}-\frac{\varphi'}{2}\right)=\sin\frac{\varphi'}{2}\,.
\end{equation}
Using (\ref{eq:eccentricity}) for the eccentricity $\epsilon$ leads to
\begin{eqnarray}\label{eq:keplerdeflection1}
\sin\frac{\varphi'}{2}=-\frac{1}{\epsilon}&=&-\left[1+E'\frac{2l^2}{\mu
    k^2}\right]^{-1/2}\nonumber\\
&=&-\left[1+E'\frac{2 \mu^2v_0'^2b^2}{\mu k^2}\right]^{-1/2}
=-\left[1+4E'^2\frac{b^2}{k^2}\right]^{-1/2}\nonumber\\
&=&
-\left[1+\frac{b^2}{b_0^2}\right]^{-1/2}\quad{\rm with}\quad b_0:=\frac{k}{2E'}\,.
\end{eqnarray}
This can be further simplified into
\begin{equation}\label{eq:keplerdeflection2}
\tan\frac{\varphi'}{2}=-\frac{b_0}{b}\,,
\end{equation}
correctly reflecting the sign convention. The scattering angle $\varphi'$ is
therefore a simple function of the normalized impact parameter $b/b_0$, where
$b_0$ has the dimension of a length:

\begin{figure*}[h]
\centerline{\includegraphics{rutherford1.eps}}
\end{figure*}

\clearpage
\subsection{Scattering cross section}
Scattering problems are often described in terms of a scattering
cross section. The scattering cross section $\sigma$ is the effective area
an object exposes to projectiles targeting the object from a particular
direction. In the simplest form, $\sigma$ is just  the projection of the
object onto a screen $S$ perpendicular to the 
velocity vector $\vec{v}$ of the projectiles before the impact:

\begin{figure}[h]
\centerline{\includegraphics{crosssection1.eps}}
\end{figure}

If the density of the projectiles per screen area is constant, the number of
projectiles hitting the object is proportional to its scattering cross section
$\sigma$.

In the previous section, we found a relation between the impact parameter $b$
and the deflection angle $\varphi'$ for scattering trajectories of two
particles that interact via a potential $U(r)$. To connect these trajectories
with the concept of a scattering cross section, we first quantify the
direction in which projectiles are scattered.  A single direction in the
3-dimensional space can be described by two angles $\theta, \phi$ in a
spherical coordinate system. The 
{\em \href{https://en.wikipedia.org/wiki/Solid_angle}{solid angle}} $\Omega$ captures a set of directions in space, and is
just the surface area on a unit sphere corresponding to this set of
directions. Since the surface of a sphere is $4\pi r^2$, the full solid angle
corresponding to all possible directions is $\Omega=4\pi$, a half space
corresponds to $\Omega=2\pi$. The solid angle is dimensionless, but
occasionally, the unit sr (for steradian) is used to indicate that reference
is made to a solid angle. \vspace{3mm}

\raisebox{-3cm}{
\includegraphics{solidangle2.eps}}\hfill
\begin{minipage}{6cm}
In spherical coordinates, a small set of directions is given by the solid angle element
\begin{equation}
d\Omega=\sin\theta\,d\phi\,d\theta\,.
\end{equation}
\end{minipage}

Coming back to scattering orbits in two-body interactions, we need to consider
two different coordinate systems. One is the coordinate system in which we
describe the two-body interaction. As we have seen in section
\ref{ssec:planemotion}, the two-body interaction is a problem in a plane $W$,
characterized by polar coordinates $\theta,r$ in the figure below:

\begin{figure}[h]
\centerline{\includegraphics{crosssection2.eps}}
\end{figure}

Remember that this
is an effective one-body problem, assuming a description in an inertial
reference frame, with the coordinate origin $O$ centered in the center-of-mass
motion of the two-body problem. The angular momentum $\vec{L}$ is constant
in time, and perpendicular to the plane $W$ of orbital motion.

The other coordinate system is used to describe the deflection of the particle
in a coordinate system suitable to describe directions. This is a spherical
coordinate system, with the main axis pointing in the direction of the
velocity $\vec{v}'_0$ of the effective single body scattered from the
origin $O$. This is a spherical coordinate system, where $z$ axis is aligned
parallel to the initial velocity $\vec{v}_0'$ of the projectile.

We consider the surface element $d\sigma$ in the screen plane
$S$ that corresponds to a deflection into the direction element $d\Omega$. The ratio
\begin{equation}
\frac{d\sigma}{d\Omega}
\end{equation}
is referred to as the {\em differential cross section}. To calculate
this quantity, we first note that the scattering problem for central force
interactions is rotationally symmetric around the $z$ axis. Projectiles hitting in a ring of radius $b$ and thickness
$db$ with an area
\begin{equation}\label{eq:differentialsigma1}
d\sigma=2\pi b\,db
\end{equation}
will have the same differential cross section, and end up in a ring-shaped
set of directions as shown in the figure below, covering a solid angle of
\begin{equation}
d\Omega=2\pi\sin\varphi'\,d\varphi'\,.
\end{equation}

\begin{figure}[h]
\centerline{\includegraphics{crosssection3.eps}}
\end{figure}

With this, we find for the differential scattering cross section
\begin{equation}\label{eq:differentialsigma}
\frac{d\sigma}{d\Omega}=\frac{2\pi b\,db}{2\pi\sin\varphi'\,d\varphi'}=
\frac{b}{\sin\varphi'}\,\left|\frac{db}{d\varphi'}\right|\,.
\end{equation}
The modulus of the derivative $db/d\varphi'$ was taken because for
scattering problems with a repulsive potential, $db/d\varphi'$ is negative;
the differential cross section is only meaningful for positive values.

\subsubsection{Rutherford scattering}
An important problem that helped to discover the structure of atoms
was the elastic scattering of $\alpha$ particles (positively charged helium
nuclei) by the also positively charged nuclei of heavy atoms.
To treat this in the framework of classical mechanics, we recall the
expression (\ref{eq:keplerdeflection1}) for the deflection angle $\varphi'$
for scattering orbits in Kepler problems in the previous section:
\begin{equation}
\sin\frac{\varphi'}{2}=-\left[1+\frac{b^2}{b_0^2}\right]^{-1/2}
\end{equation}
By squaring and inverting this expression, we get
\begin{equation}
\left(\sin\frac{\varphi'}{2}\right)^{-2}=1+\frac{b^2}{b_0^2}\,,
\end{equation}
which can be differentiated on both sides,
\begin{equation}
-2\sin^{-3}\frac{\varphi'}{2}\cos\frac{\varphi'}{2}\,\frac{1}{2}\,d\varphi'=\frac{2\,b}{b_0^2}\,db\,,
\end{equation}
or
\begin{equation}
\frac{db}{d\varphi'}=-\frac{b_0^2\cos(\varphi'/2)}{\sin^3(\varphi'/2)}\,\frac{1}{2b}\,,
\end{equation}
which is negative for all scattering angles $0\le\varphi'\le\pi$. We use this
result in expression (\ref{eq:differentialsigma}) for the differential
scattering cross section, and obtain
\begin{eqnarray}
\frac{d\sigma}{d\Omega}&=&\frac{b_0^2\cos(\varphi'/2)}{2\sin\varphi'\sin^3(\varphi'/2)}\nonumber\\
%&=&\frac{b_0^2\cos(\varphi'/2)}{2\cdot2\cos(\varphi'/2)\sin(\varphi'/2)\sin^3(\varphi'/2)}\nonumber\\
&=&\frac{b_0^2}{4\sin^4(\varphi'/2)}=\frac{k^2}{16E'^2\sin^4(\varphi'/2)}\,.
\end{eqnarray}
This is a historically and practically important formula for {\em
  \href{https://en.wikipedia.org/wiki/Rutherford_scattering}{Rutherford
    scattering}}. The fortunate fact that this result,
obtained by classical mechanics, is the same as the one obtained from a
quantum mechanical description led to a rapid development of the understanding
of the structure of atoms by
Rutherford\footnote{published in Philosophical Magazine {\bf 21}, 669-688
  (1911) by
  \href{https://en.wikipedia.org/wiki/Ernest_Rutherford}{Ernest Rutherford},
  1871-1937 } and coworkers.
\subsection{Scattering problems in the laboratory system}
The differential scattering cross section $d\sigma/d\Omega$ is very useful
because it is directly related to typical scattering experiments. Usually, a 
scattering target is located somewhere in space, and exposed to a homogenous
flux of projectile particles (e.g. Helium nuclei or protons). A typical sample 
consists of many target particles, and is much larger than a
characteristic impact parameter for an individual scattering process, like
$b_0$ in (\ref{eq:keplerdeflection1}). A
detector with a fixed size measures the flux of scattered particles at
different deflection angles in a fixed distance from the target, so the solid
angle subtended by the detector from the view of the scattering target is
fixed. The  differential scattering cross section is than simply proportional
to the detected number of scattered particles in a particular direction.

To make a connection between the differential scattering cross section seen in
the lab with the one calculated for the
effective one-body problem, we need to find an adequate transformation --
first into the centre-of-mass (CM) system, then into the lab.

\begin{figure}[h]
\centerline{\includegraphics{scattering3.eps}}
\end{figure}

The center-of-mass position $\vec{R}$ of two masses $m_1, m_2$ at positions
$\vec{r}_1$ and $\vec{r}_2$ (in the lab system) was was defined in section \ref{ssec:COMmotion} as
\begin{equation}\label{eq:COMposition}
\vec{R}=\frac{1}{m_1+m_2}\left(m_1\vec{r}_1+m_2\vec{r}_2\right)=\frac{\mu}{m_2}\vec{r}_1+\frac{\mu}{m_1}\vec{r}_2\,,
\end{equation}
with the effective mass $\mu=m_1m_2/(m_1+m_2)$. The coordinate origin in a CM
system is chosen such that $\vec{R}=0$. Then, the
positions $\vec{r}_1'$ and $\vec{r}_2'$ of both masses in the CM
system are connected by
\begin{equation}
m_1\,\vec{r}'_1+m_2\,\vec{r}'_2=0\,,\qquad{\rm or}\qquad
\vec{r}_1'=\frac{\mu}{m_1}\,\vec{r}\,,\quad
\vec{r}_2'=-\frac{\mu}{m_2}\,\vec{r}
\end{equation}
with the difference vector
\begin{equation}\label{eq:differencevector}
\vec{r}=\vec{r}_1'-\vec{r}_2' = \vec{r}_1-\vec{r}_2\,,
\end{equation} describing completely the state of the system in the effective
single-particle description. With (\ref{eq:COMposition}) and
(\ref{eq:differencevector}), the positions of particles 1 and 2 in the lab
system can be expressed by $\vec{R}$ and $\vec{r}$ according to
\begin{equation}
\vec{r}_1=\vec{R}+\vec{r}_1'=\vec{R}+\frac{\mu}{m_1}\,\vec{r}\,,
\quad{\rm and}\quad
\vec{r}_2=\vec{R}-\frac{\mu}{m_2}\,\vec{r}\,.
\end{equation}
Similarly, the velocities in the lab system are given by
\begin{equation}\label{eq:labvelocities}
\dot\vec{r}_1=\dot\vec{R}+\frac{\mu}{m_1}\,\dot\vec{r}
\quad{\rm and}\quad
\dot\vec{r}_2=\dot\vec{R}-\frac{\mu}{m_2}\,\dot\vec{r}\,.
\end{equation}

\subsubsection{Transformation of energy between lab and CM system}
The total kinetic energy of the two particles in the lab system at any time
can be expressed by the CM velocity $\dot\vec{R}$ and the effective
single-particle velocity $\dot\vec{r}$,
\begin{eqnarray}
T=\frac{1}{2}m_1\,\dot\vec{r}_1^2+\frac{1}{2}m_2\,\dot\vec{r}_2^2&=&
\frac{1}{2}m_1\left(\dot\vec{R}+\frac{\mu}{m_1}\,\dot\vec{r}\right)^2+
\frac{1}{2}m_1\left(\dot\vec{R}-\frac{\mu}{m_2}\,\dot\vec{r}\right)^2\nonumber\\
&=&\frac{1}{2}(m_1+m_2)\,\dot\vec{R}^2+\frac{1}{2}\left(\frac{\mu^2}{m_1}+\frac{\mu^2}{m_2}\right)\,\dot\vec{r}^2\nonumber\\
&=&\frac{1}{2}M\dot\vec{R}^2+\frac{1}{2}\mu\,\dot\vec{r}^2=\frac{1}{2}M\dot\vec{R}^2+T'
\end{eqnarray}
with the total mass $M=m_1+m_2$, and the kinetic energy $T'$ in the CM
system. Long before and long after the impact, all the
energy is in the kinetic energy of the particles, so
\begin{equation}\label{eq:energyrelation1}
E=T=\frac{1}{2}M\dot\vec{R}^2+E'\,,
\end{equation}
where $E'$ is the energy in the CM system available for the scattering
process.

In a typical scattering scenario, the target mass $m_2$ is
initially at rest in the lab (i.e., $\dot\vec{r}_2=0$), and the projectile
mass $m_1$ moves towards the target with a velocity $\vec{v}_0$. By
differentiating (\ref{eq:COMposition}), one finds for the CM velocity
\begin{equation}\label{eq:COMvelocity}
\dot\vec{R}=\frac{\mu}{m_2}\vec{v}_0\,.
\end{equation}
By using (\ref{eq:labvelocities}), one finds
\begin{equation}
\vec{v}_0=\dot\vec{r}_1=\frac{\mu}{m_2}\vec{v}_0+\frac{\mu}{m_1}\dot\vec{r}
\quad{\rm or}\quad \vec{v}_0=\dot\vec{r}=\vec{v}'_0\,,
\end{equation}
i.e., the velocity $\vec{v}'_0$ in the CM system before the interaction is
the same as the initial velocity $\vec{v}_0$ of the projectile in the lab
system. Acknowledging
that the initial total energy is only given by the kinetic energy
$m_1\vec{v}_0^2/2$ of the projectile, and using (\ref{eq:COMvelocity}) in
(\ref{eq:energyrelation1}), one finds the simple relation
\begin{equation}\label{eq:energytransfomration}
E'=E\frac{m_2}{m_1+m_2}
\end{equation}
for the total available energy $E'$ in the CM system, and therefore also in
the effective single-particle system. Since the impact
parameter $b$ in the lab, the CM system, and in the effective single particle
system are the same, we  have enough information
to calculate the deflection angle $\varphi'=\varphi'(E',b)$ (in the effective
single particle systm) from a known energy $E$ in the lab system. Also note
that the angle $\varphi'$ between $\vec{r}$ and $\vec{v}_0$ is the same in the
effective single particle system and the CM system.

\subsubsection{Transformation of deflection angles between lab and CM system}
To transform the deflection angle $\varphi'$ back to the lab system, we
consider the geometrical relationship between velocities
$\vec{r}_1$, $\vec{r}_2$ and $\vec{r}$ as shown:

\begin{figure}[h]
\centerline{\includegraphics{scattering4.eps}}
\end{figure}
On the left side, the relationship between $\varphi$ and the asymptotic
trajectories in the lab, and the deflection angle $\varphi'$ with respect to
the difference vector $\vec{r}$ is shown. On the right side, one can see the
corresponding geometry of the velocity relation
$\dot\vec{r}_1=\dot\vec{R}+\dot\vec{r}'_1$ between lab and CM
system. Splitting this up into Cartesian components leads to
\begin{eqnarray}
\dot{r}_1\cos\varphi&=&\dot{r}'_1\cos\varphi'+\dot{R}\quad{\rm and}\nonumber\\
\dot{r}_1\sin\varphi&=&\dot{r}'_1\sin\varphi'\,.
\end{eqnarray}
Note that this expression relates asymptotic velocities,
so $\varphi$ and $\varphi'$ do not change with time anymore.
This can be resolved into a relation between $\varphi$ and $\varphi'$,
\begin{equation}\label{eq:angletranslation}
\tan\varphi=\frac{\sin\varphi}{\cos\varphi}=\frac{\sin\varphi'}{\cos\varphi'+\gamma}\,,\quad{\rm
  with}\quad\gamma:=\frac{\dot{R}}{\dot{r}'_1}=\frac{m_1}{m_1+m_2}\,\frac{v_0}{\dot{r}'_1}\,,
\end{equation}
which still depends on the final speed $\dot{r}'_1$ of the projectile in the
CM system. To eliminate this, we express the total energy $E'=T'$ in the CM
system {\em before} the impact by the kinetic energies {\em after} the impact,
and include a possible energy loss $H$ during the scattering process:
\begin{equation}
E'=\frac{1}{2}m_1\dot{r}'^2_1+\frac{1}{2}m_2\dot{r}'^2_2+H\,.
\end{equation}
This can e.g. cover a radiation losses due to the
acceleration of charges in the scattering process. The kinetic 
energy of $m_2$  can also be expressed by $\dot{r}'_1$
using momentum conservation in the CM system, leading to
\begin{equation}
E'=\frac{1}{2}m_1\left(\frac{m_1+m_2}{m_2}\right)\dot{r}'^2_1+H\,.
\end{equation}
By expressing $E'$ by $E$ via (\ref{eq:energytransfomration}) and dividing by
$E$, one finds
\begin{equation}
\frac{m_2}{m_1+m_2}=\frac{1}{E}\frac{1}{2}m_1\left(\frac{m_1+m_2}{m_2}\right)\dot{r}'^2_1+\frac{H}{E}
\end{equation}
or
\begin{equation}
1=\underbrace{\left(\frac{m_1+m_2}{m_2}\right)^2\frac{\dot{r}'^2_1}{v_0^2}}_{=\frac{1}{\gamma^2}\frac{m_1^2}{m_2^2}}+\frac{H}{E}\frac{m_1+m_2}{m_2}
\end{equation}
which can finally be resolved into 
\begin{equation}
\gamma^2=\left(\frac{m_1}{m_2}\right)^2\frac{1}{1-\frac{H}{E}\left(\frac{m_1+m_2}{m_2}\right)}\,.
\end{equation}
For elastic scattering with $H=0$, one gets the simple relation
\begin{equation}
\gamma=\frac{m_1}{m_2}
\end{equation}
for use in the transformation (\ref{eq:angletranslation}) between deflection
angles $\varphi$ and $\varphi'$.

\subsubsection{Transformation of the differential scattering cross sections}
From the dependence of the deflection angle $\varphi'$ on the impact parameter
$b$, the differential scattering cross section $d\sigma/d\Omega$ could be
calculated via (\ref{eq:differentialsigma}). Since the differential
$d\sigma=2\pi b\,db$ in (\ref{eq:differentialsigma1}) is the same in the lab and 
CM system, we need to change the expression for solid angle element $d\Omega$
by one for the lab system:
\begin{equation}
d\Omega_{\rm Lab}=2\pi\sin\varphi\,d\varphi\,,
\end{equation}
which leads to the simple relation between the differential scattering cross
sections in the lab and the CM system,
\begin{equation}\label{eq:diffsigmatransformation}
\left(\frac{d\sigma}{d\Omega}\right)_{\rm Lab}=\left(\frac{d\sigma}{d\Omega}\right)_{\rm CM}\,\frac{\sin\varphi'}{\sin\varphi}\frac{d\varphi'}{d\varphi}\,,
\end{equation}
where the differential scattering cross sections are evaluated at
corresponding angles $\varphi$ in the lab, and $\varphi'$ in the CM system.
%
Evaluation of the correction factor on the right side of
(\ref{eq:diffsigmatransformation}) is straightforward but a bit tedious; using
(\ref{eq:angletranslation}), i.e.,
\begin{equation}
\tan\varphi=\frac{\sin\varphi'}{\cos\varphi'+\gamma}
\end{equation}
one finds with $1/\tan^2 x=1/\sin^2x-1$ first
\begin{equation}
\frac{\sin\varphi'}{\sin\varphi}=\sqrt{1+\gamma^2+2\gamma\cos\varphi'}\,.
\end{equation}
Expressing the differential of (\ref{eq:angletranslation}) on the left side by
$d\varphi$, and on the right side by $d\varphi'$ leads after some steps to
\begin{equation}
\frac{d\varphi'}{d\varphi}=\frac{1+\gamma^2+2\gamma\cos\varphi'}{1+\gamma\cos\varphi'} \,,
\end{equation}
and finally to a correction factor
\begin{equation}\label{eq:diffsigmacorrection}
\frac{\sin\varphi'}{\sin\varphi}\frac{d\varphi'}{d\varphi}=
\frac{(1+\gamma^2+2\gamma\cos\varphi')^{3/2}}{1+\gamma\cos\varphi'}
\end{equation}

\subsubsection{Special cases}
For the special case of elastic scattering ($H=0$), and $m_1=m_2$, one finds
\begin{equation}
\gamma=1 \quad{\rm and}\quad E'=E/2\,,
\end{equation}
so
\begin{equation}
\tan\varphi=\frac{\sin\varphi'}{\cos\varphi'+1}=\tan\frac{\varphi}{2}\,,\quad{\rm or}\quad\varphi=\frac{\varphi'}{2}\,.
\end{equation}
The correction factor (\ref{eq:diffsigmacorrection}) becomes
\begin{equation}
\frac{\sin\varphi'}{\sin\varphi}\frac{d\varphi'}{d\varphi}=4\cos\varphi\,,
\end{equation}
and
\begin{equation}
\left(\frac{d\sigma}{d\Omega}\right)_{\rm Lab}=\left(\frac{d\sigma}{d\Omega}\right)_{{\rm CM}\,,\varphi'=2\varphi}\cdot4\cos\varphi\,.
\end{equation}
This case covers e.g. scattering of electrons electrons, or protons by
protons.

The other important special case of elastic scattering ($H=0$) is for a
target mass $m_2$ much larger than the projectile mass $m_1$. Then,
$\gamma\approx0$ and
\begin{equation}
\varphi'\approx\varphi\,,\quad E'=E\,,\quad{\rm and}\quad
\left(\frac{d\sigma}{d\Omega}\right)_{\rm Lab}=\left(\frac{d\sigma}{d\Omega}\right)_{{\rm CM}\,,\varphi'=\varphi}\,.
\end{equation}
This was the case for the classical {\em \href{https://en.wikipedia.org/wiki/Geiger-Marsden_experiment}{Rutherford scattering experiment}}
conducted by Geiger and Marsden\footnote{H. Geiger and E. Marsden,
  \href{https://dx.doi.org/10.1098/rspa.1909.0054}{Proc. Roy. Soc. London A{\bf 82}, 495-500 (1909)}.}, where
relatively light $\alpha$ particles ($m_1=4$\,amu) were scattered of a thin
foil of gold atoms ($m_2\approx197$\,amu). The occasional relatively rare
scattering of $\alpha$ particles in the backwards direction
($\varphi>90^\circ$) indicated that the positive charge of the nuclei was
localized in a very small space only, and not uniformly distributed over the
whole size of the atom, as hypothesized by the then common ``plum pudding''
model of a large-sized positive charge to counterbalance the electrons.

\clearpage
\section{Harmonic oscillator}
So far, we have encountered the undamped harmonic oscillator as an
example for obtaining an equation of motion via various strategies, resulting in
\begin{equation}
F=m\ddot{x}=-kx\quad{\rm or}\quad \ddot{x}+\omega_0^2x=0
\end{equation}
with $\omega_0^2=k/m$ for the dynamical variable $x$, with oscillating
solutions discussed in 
section \ref{sssec:harmonicosc1}. Since the harmonic oscillator is at the core
of many dynamic phenomena in physics, it deserves a closer look, and the
inclusion of dissipation as well as response to time-varying external forces.
\subsection{Damped harmonic oscillator}
First, we extend the basic model by a damping term. This can be thought of as
a friction force present in the motion. In its simplest form (see section
\ref{ssec:friction}), the friction force would be proportional to the velocity
$\dot{x}$ of the system. This results in a modified equation of motion,
\begin{equation}
F=m\ddot{x}=-kx-\alpha\dot{x}\,,
\end{equation}
or in the more commonly found form
\begin{equation}\label{eq:dampedshoeom}
\ddot{x}+2\beta\dot{x}+\omega_0^2x=0\,,
\end{equation}
with $\beta>0$ for a damping action.
This is still a linear ordinary differential equation (ODE), and can also be
solved using an exponential ansatz
\begin{equation}\label{eq:dampedshoansatz}
x(t)=Ae^{rt}\,.
\end{equation}
Inserting this into (\ref{eq:dampedshoeom}) and division by $x(t)$
leads to an algebraic equation for $r$,
\begin{equation}\label{eq:dampedshoalgebraic}
r^2+2\beta r+\omega_0^2=0\,,
\end{equation}
which can easily be solved for its roots:
\begin{equation}\label{eq:dampedshoroots}
r_{1,2}=-\beta\pm\sqrt{\beta^2-\omega_0^2}\,.
\end{equation}
Depending on the values of $\omega_0$ and $\beta$, the two roots $r_{1,2}$ can
assume complex values. One distinguishes three cases for the solutions.

\subsubsection{Underdamped case}
If $0<\beta<\omega_0$, the argument of the square root in
(\ref{eq:dampedshoroots}) is negative, so $r$ has an imaginary component:
\begin{equation}
r_{1,2}=-\beta\pm i\sqrt{\omega_0^2-\beta^2}=-\beta\pm i\omega_1
\quad{\rm with}\quad \omega_1:=\sqrt{\omega_0^2-\beta^2}\,.
\end{equation}
The corresponding solutions (\ref{eq:dampedshoansatz}) are
\begin{equation}
x(t)=Ae^{-\beta t \pm i\omega_1 t}=Ae^{-\beta t}e^{\pm i\omega_1t}\,.
\end{equation}
The first exponential provides the damping term that decays exponentially with
time, while the second exponential forms an oscillating part, with an
oscillation frequency $\omega_1$ lower than the frequency $\omega_0$
of the undamped system. As this is
a second order differential equation, there are two integration constants
that allow meeting the initial conditions of the problem; the
most general solution to (\ref{eq:dampedshoeom}) can be written in various ways,
\begin{eqnarray}\label{eq:dampedshosolution}
x(t)&=&(Ae^{i\omega_1t}\,+\,A'e^{-i\omega_1t})e^{-\beta t}\nonumber\\
&=&(B\cos\omega_1t\,+\,B'\sin\omega_1t)e^{-\beta t}\nonumber\\
&=&C\cos(\omega_1t-\delta)e^{-\beta t}
\end{eqnarray}
for integration constant pairs $(A,A')$, $(B,B')$, or $(C,\delta)$. The first
form is convenient if complex parameters are to be considered, the second and
third are often useful when a real-valued $x(t)$ is expected. 
The solutions are illustrated below, where an
oscillation of frequency $\omega_1<\omega_0$ is multiplied with an
exponentially decaying {\em envelope} $e^{-\beta t}$. For
$t\rightarrow\infty$, the exponential term takes over, and $x(t)\rightarrow0$.


\begin{figure*}[h]
\centerline{\includegraphics{dampedoscillation.eps}}
\end{figure*}


\subsubsection{Critically damped case}
For $\beta=\omega_0$, the two roots of (\ref{eq:dampedshoalgebraic}) become
degenerate, $r_{1,2}=-\beta$. In this case, one can verify that the ansatz
\begin{equation}\label{eq:dampedshocriticalsolution}
x(t)=(A+Bt)\,e^{-\beta t}
\end{equation}
is a solution to (\ref{eq:dampedshoeom}), which can cover all initial
conditions. Typical solutions for $x(t)$ matching various initial conditions
are shown below:

\begin{figure*}[h]
\centerline{\includegraphics{criticallydampedosc.eps}}
\end{figure*}

\subsubsection{Overdamped case}
For $\beta>\omega_0$, both roots $r_{1,2}$ in (\ref{eq:dampedshoroots}) are
real and negative, because the square root 
\begin{equation}
\sqrt{\beta^2-\omega_0^2}=:\omega_2
\end{equation}
is always smaller than $\beta$. The solution for $x(t)$ is
a superposition of two exponential decays, with two time constants
corresponding to $r_1$ and $r_2$:
\begin{eqnarray}\label{eq:overdampedshosolution}
x(t)&=&(Ae^{\omega_2t}+A'e^{-\omega_2t})e^{-\beta t}\nonumber\\
&=&Be^{-(\beta+\omega_2)t}+B'e^{-(\beta-\omega_2)t}\,.
\end{eqnarray}
The figure below shows a few typical examples for $x(t)$:

\begin{figure*}[h]
\centerline{\includegraphics{overdamped.eps}}
\end{figure*}

\subsubsection{Damped oscillator trajectories in phase space}
It is instructive to visualize the various cases for solutions to the harmonic
oscillator problem in phase space. As introduced in section
\ref{ssec:phasespace}, the phase space for the simple harmonic oscillator is a
plane with coordinates $x$ and $p=mv=m\dot{x}$, i.e., the momentum is
proportional to the speed of the oscillator.

On the left side of the figure below, the case $\beta=0$ reproduces the closed
elliptical trajectories we have seen in section \ref{ssec:phasespace}. Various
initial conditions lead to trajectories with different amplitude.

\begin{figure*}[h]
\centerline{\includegraphics{phasespacessho.eps}}
\end{figure*}

In the center, the underdamped case is illustrated for a single initial
condition with a given amplitude and velocity. the trajectory is a logarithmic
spiral, converging into the origin $x=0,p=0$ for $t\rightarrow\infty$.

On the right side, a few trajectories for the overdamped case are shown, with a
fixed value $\beta=1.3\omega_0$. The two dashed lines correspond to solutions
of type 
(\ref{eq:overdampedshosolution}) with either $B=0$, or $B'=0$ (fast decay or
slow decay only). There, the ratio between position $x$ and velocity $\dot{x}$
is fixed and given by the respective decay rates $-\beta-\omega_2$ and
$-\beta+\omega_2$. The trajectories first follow a direction corresponding to
the faster decay rate $-\beta-\omega_2$. After some time, this contribution has
become much smaller 
than the one with the slower decay rate $-\beta+\omega_2$, which then dominates
how the trajectory approaches the origin of the phase space.


\subsection{Harmonically driven harmonic oscillator}
So far, we have only considered time-independent equations of motion,
which are adequate for closed systems. However, physical systems are often
driven by external forces, so it is important to know how the system
responds. Staying with the (damped) harmonic oscillator, we first consider an
external force $F_{\rm ext}(t)=F_0\cos(\omega t)$ with a harmonic 
time dependence:
\begin{equation}\label{eq:drivenosc1}
\ddot{x}+2\beta\dot{x}+\omega_0^2x=A\cos(\omega t)
\quad{\rm with}\quad A=\frac{F_0}{m}\,.
\end{equation}
This is a so-called {\em inhomogeneous} linear differential equation, with the
inhomogeneity on the right side of the equation. The solution $x(t)$ of such
a differential equation is the sum of a {\em complementary solution}
$x_c(t)$ to the homogenous differential equation, as it was earlier presented in
(\ref{eq:dampedshosolution}), (\ref{eq:dampedshocriticalsolution}), or
(\ref{eq:overdampedshosolution}), and a {\em particular solution} $x_p(t)$ that
takes care of the sinusoidal driving part. We first solve this problem
with a real-valued ansatz, and later with a complex-valued one. The first
approach guarantees a real-valued solution, but the latter is
algebraically simpler, and easier to derive and remember.

\subsubsection{Solution with a real-valued ansatz}
The inhomogeneity oscillates with a fixed (angular) frequency $\omega$,
so we expect a periodic solution $x(t)$ with the
same angular frequency. Similar to the ansatz in section
\ref{sssec:harmonicosc1}, we allow for a phase shift $\delta$ and try
\begin{equation}
x(t)=B\cos(\omega t-\delta)\,.
\end{equation}
Direct differentiation and inserting into (\ref{eq:drivenosc1}) leads to
\begin{equation}
-B\omega^2\cos(\omega t-\delta)-2\beta B\omega\sin(\omega
t-\delta)+\omega_0^2B\cos(\omega t-\delta)=A\cos(\omega t)\,.
\end{equation}
This can be sorted into terms with $\sin$ and $\cos$ and rearranged:
\begin{equation}\label{eq:drivenosc2}
(\omega_0^2-\omega^2)\,\cos(\omega t-\delta)\,-\,2\beta\omega\,\sin(\omega
t-\delta)=\frac{A}{B}\cos(\omega t)\,.
\end{equation}
The oscillating terms with the phase shift $\delta$ on the left side can
be expanded into an oscillating and a static part:
\begin{eqnarray}
\cos(\omega t-\delta)&=&\cos\omega t\,\cos\delta+\sin\omega
t\,\sin\delta\nonumber\\
\sin(\omega t-\delta)&=&\sin\omega t\,\cos\delta-\cos\omega
t\,\sin\delta
\end{eqnarray}
With this, (\ref{eq:drivenosc2}) can be arranged in a terms proportional
to $\sin\omega t$ and $\cos\omega t$:
\begin{eqnarray}
\left[(\omega_0^2-\omega^2)\cos\delta+2\beta\omega\sin\delta\right]\cos\omega
t&&\nonumber\\
+\left[(\omega_0^2-\omega^2)\sin\delta-2\beta\omega\cos\delta\right]\sin\omega
t&=&\frac{A}{B}\cos\omega t\,.
\end{eqnarray}
Since this equation needs to hold for all times, the amplitude of the $\sin$
and $\cos$ components on both sides need to be the same, which leads to the
two equations
\begin{eqnarray}
(\omega_0^2-\omega^2)\cos\delta+2\beta\omega\sin\delta&=&\frac{A}{B}\,,\quad{\rm and}\label{eq:drivensho3}\\
(\omega_0^2-\omega^2)\sin\delta-2\beta\omega\cos\delta&=&0\,.\label{eq:drivensho3a}
\end{eqnarray}
Equation (\ref{eq:drivensho3a}) leads to an expression for the phase shift,
\begin{equation}\label{eq:realphaseshift}
\tan\delta=\frac{2\beta\omega}{\omega_0^2-\omega^2}\,,
\end{equation}
which provides expressions for $\sin\delta$ and $\cos\delta$':
\begin{equation}
\sin\delta=\frac{2\beta\omega}{\sqrt{(\omega_0^2-\omega^2)^2+4\omega^2\beta^2}}
\quad{\rm and}\quad
\cos\delta=\frac{\omega_0^2-\omega^2}{\sqrt{(\omega_0^2-\omega^2)^2+4\omega^2\beta^2}}\,.
\end{equation}
Those can be used to find a relationship between the amplitudes $A$ of the
driving acceleration, and the amplitude $B$ of the resulting oscillation from
(\ref{eq:drivensho3}):
\begin{equation}\label{eq:suscept1}
\frac{B}{A}=\frac{1}{(\omega_0^2-\omega^2)\cos\delta+2\beta\omega\sin\delta}=
\frac{1}{\sqrt{(\omega_0^2-\omega^2)^2+4\omega^2\beta^2}}
\end{equation}
Before we discuss the result, we show that by using complex amplitudes, the
problem can be substantially simplified, without using relations for the
trigonometric functions.

\subsubsection{Solution with a complex-valued driving term}
We first re-write the inhomogeneous differential equation (\ref{eq:drivenosc1})
into  one with a complex-valued inhomogeneity:
\begin{equation}\label{eq:drivenosc4}
\ddot{x}+2\beta\dot{x}+\omega_0^2x=Ae^{i\omega t}\,.
\end{equation}
Inserting the complex-valued ansatz
\begin{equation}\label{eq:drivenosc5}
x_p(t)=Be^{i\omega t}
\end{equation}
into the differential equation (\ref{eq:drivenosc4}) gives
\begin{equation}\label{eq:drivenosc6}
-\omega^2Be^{i\omega t}\,+\,2i\beta\omega Be^{i\omega t}\,+\,\omega_0^2Be^{i\omega t}=Ae^{i\omega t}\,,
\end{equation}
leading to the algebraic equation
\begin{equation}\label{eq:drivenosc7}
-\omega^2+2i\beta\omega+\omega_0^2=\frac{A}{B}
\end{equation}
or
\begin{equation}\label{eq:complexsusceptibility}
\frac{1}{(\omega_0^2-\omega^2)+2i\beta\omega}=\frac{B}{A}=:\chi(\omega)\,.
\end{equation}
Here, we defined the {\em complex susceptibility} $\chi(\omega)$ which
specifies the ratio between the amplitude $B$ of the oscillator response, and
the amplitude $A$ of the driving oscillation. The modulus of the complex
susceptibility is easy to calculate,
\begin{equation}\label{eq:modsusceptibility}
\left|\chi(\omega)\right|=\frac{1}{\left|(\omega_0^2-\omega^2)+2i\beta\omega\right|}=
\frac{1}{\sqrt{(\omega_0^2-\omega^2)^2+4\beta^2\omega^2}}
\end{equation}
and reproduces the result (\ref{eq:suscept1}) for the amplitude ratio for a
system driven by a real-valued harmonic inhomogeneity. Similarly, the {\em
  argument} ${\rm arg}(\chi)$ of the complex susceptibility reflects the phase
shift between  complex amplitudes $B$ and $A$. For this, we remember that a
complex number $z$ can be written as
\begin{equation}
z=|z|e^{i\,{\rm arg}(z)}\quad{\rm with}\quad \tan({\rm arg}(z))=\frac{{\rm
    Im}[z]}{{\rm Re}[z]}
\end{equation}
With $1/(a+ib)=(a-ib)/(a^2+b^2)$, we find from (\ref{eq:complexsusceptibility})

\begin{equation}
\tan({\rm arg}(\chi))=\frac{-2\beta\omega}{\omega_0^2-\omega^2}=\tan(-\delta)\,,
\end{equation}
which reproduces the phase shift $\delta$ for the real-valued expression (\ref
{eq:realphaseshift}). The minus sign in the phase shift $\delta$ reflects the
fact that we can write
\begin{equation}
x_p(t)=\chi(\omega)Ae^{i\omega t}=A|\chi(\omega)|e^{-i\delta}e^{i\omega
  t}=A|\chi(\omega)|e^{i(\omega t-\delta)}\,,
\end{equation}
where a positive valued $\delta$ reflects a {\em phase lag} of the response
with respect to the driving acceleration.

\subsubsection{Transients in harmonically driven oscillators}
As mentioned before, the solution $x(t)$ to the harmonically driven oscillator
is a linear combination of the particular solution we just obtained, and the
complementary solution to the homogenous differential equation:
\begin{equation}
x(t)=x_c(t)+x_p(t)\,.
\end{equation}
The complementary solution allows taking care of the initial conditions of the
system, because the particular solution leaves no freedom to do so. For a time $t$ long after the instant where the initial conditions
were defined, the complementary solution $x_c(t)$ will have decayed for the
damped harmonic oscillator, and the particular solution $x_p(t)$ will dominate
the response. The complementary solution leads to a so-called {\em
  transient}, shown in the figure below for two examples:

\begin{figure*}[h]
\centerline{\includegraphics{transients.eps}}
\end{figure*}

\subsubsection{Stationary solution of the driven harmonic oscillator}
After the transients of the solution $x(t)$ are gone, the system assumes the
stationary solution
$x_p(t)$, which is, apart from an amplitude, completely characterized  by the
complex susceptibility $\chi(\omega)$ from (\ref{eq:complexsusceptibility}). 
With
\begin{equation}
\chi(\omega=0)=\frac{1}{\omega_0^2}=:\chi_0\,.
\end{equation}
and another customary definition of the so-called {\em quality factor} 
\begin{equation}
Q:=\frac{\omega_0}{2\beta}
\end{equation}
that normalizes the damping factor $\beta$ to the resonance frequency
$\omega_0$ of the undamped oscillator, the behavior of the susceptibility for
different frequencies can then be discussed in a generic way. The modulus
of the  susceptibility, normalized  to $\chi_0$ and the phase shift
$\delta=-{\rm arg}[\chi(\omega)]$ is shown in the following figure below
for different values of $Q$.
\clearpage

\begin{figure*}[h]
\centerline{\includegraphics{susceptibility.eps}}
\end{figure*}

The modulus $|\chi(\omega)|$ rises from $\chi_0$ with increasing frequency to a
single maximum at the {\em resonance frequency} $\omega=\omega_R$. The
maximum is found by differentiating (\ref{eq:modsusceptibility}),
\begin{eqnarray}
\frac{\partial|\chi(\omega)|}{\partial\omega}&=&\frac{-1/2}{\sqrt{(\omega_0^2-\omega^2)^2+4\beta^2\omega^2}^3}\left[2(\omega_0^2-\omega^2)(-2\omega)+8\omega\beta^2\right]\nonumber\\
&=&\frac{-2\omega}{\sqrt{\cdots}^3}\left[\omega^2-\omega_0^2+2\beta^2\right]
\stackrel{!}{=}0
\end{eqnarray}
revealing the maximum at
\begin{equation}
\omega=\sqrt{\omega_0^2-2\beta^2}=:\omega_R\,.
\end{equation}
This amplitude resonance frequency $\omega_R$, the damped free oscillation
frequency $\omega_1$, and the undamped oscillator frequency $\omega_0$ obey the
relation
\begin{equation}
\omega_0>\omega_1>\omega_R\,.
\end{equation}
At $\omega=\omega_0$, the phase shift $\delta$ between driving force and
oscillation amplitude has increased from 0 at $\omega=0$ to $\delta=\pi/2$,
i.e., the response of the oscillator on resonance lags a quarter period
behind to the driving force. 
The maximal value of the susceptibility modulus at the amplitude resonance
$\omega_R$ takes the value
\begin{eqnarray}
|\chi(\omega_R)|&=&\frac{1}{\sqrt{(\omega_0^2-\omega_R^2)^2+4\beta^2\omega_R^2}}\nonumber\\
&=&\frac{1}{\sqrt{4\beta^4+4\beta^2(\omega_0^2-2\beta^2)}}=\frac{1}{2\beta\sqrt{\omega_0^2-\beta^2}}=\frac{1}{2\beta\omega_1}\nonumber\\
&=&\frac{\chi_0\omega_0^2}{2\beta\omega_0\sqrt{1-\beta^2/\omega_0^2}}
=\frac{\chi_0Q}{\sqrt{1-1/(4Q^2)}}\,.
\end{eqnarray}
For a weakly damped oscillator, $Q\gg1$, so
\begin{equation}
\frac{|\chi(\omega_R)|}{\chi_0}\approx Q\,,
\end{equation}
i.e., the resonance amplitude is increased by the factor of $Q$ compared to
the low frequency response at $\omega=0$.
For $\omega\gg\omega_0$,
\begin{equation}
|\chi(\omega)|\approx\frac{1}{\omega^2}\,,
\end{equation}
and the phase lag approaches $\delta=\pi$, i.e., the response $x(t)$ of the
driven system is opposed to the driving force. The $1/\omega^2$ dependence is
e.g. used to damp out high frequency vibrations on optical tables by making
the table/suspension system an oscillator with a very low resonance frequency.


The resonance peak gets more pronounced for larger values of $Q$. To see this,
we introduce a {\em detuning} $\Delta:=\omega-\omega_0$ from the resonance,
and approximate the susceptibility for $\Delta\ll\omega_0$:
\begin{eqnarray}
|\chi(\omega)|&=&\frac{1}{\sqrt{(\omega+\omega_0)^2(\omega-\omega_0)^2+4\beta^2\omega^2}}\nonumber\\
&=&\frac{1}{\sqrt{(2\omega_0+\Delta)^2\Delta^2+\omega_0^2/Q^2(\omega_0+\Delta)^2}}\nonumber\\
&\approx&\frac{1}{\sqrt{4\omega_0^2\Delta^2+\omega_0^4/Q^2}}=
\frac{Q}{\omega_0^2\sqrt{4Q^2\Delta^2/\omega_0^2+1}}\nonumber\\
&=&\frac{\chi_0\,Q}{\sqrt{4Q^2\Delta^2/\omega_0^2+1}}
\end{eqnarray}
This is the amplitude version of a so-called Lorentz profile, which governs
resonance phenomena in many areas in physics, including spectral line of atoms
and molecules.

\begin{figure*}[h]
\centerline{\includegraphics{resonancepeak.eps}}
\end{figure*}

One can assign a width $\Delta\omega$ of the resonance, defined by the
frequency range where the average energy stored in the resonator exceeds half
of the maximal energy on resonance. Since the stored energy is proportional to
$x^2$ and therefore proportional to $|\chi^2|$, we find the condition
\begin{equation}
\left|\chi\left(\omega_0\pm\frac{\Delta\omega}{2}\right)\right|=\frac{Q}{\sqrt{2}}\quad\rightarrow\quad\Delta\omega=\frac{\omega_0}{Q}
\end{equation}
The maximum for the modulus of the susceptibility is located at
\begin{equation}
\omega_R=\sqrt{\omega_0^2-2\beta^2}=\omega_0\sqrt{1-\frac{1}{2Q^2}}\approx\omega_0-\frac{\omega_0}{4Q^2}=\omega_0-\frac{\Delta\omega}{4Q}\,,
\end{equation}
i.e., the peak separation from $\omega_0$ is much smaller than the width
$\Delta\omega$ of the resonance line for large $Q$. An example of
mechanical resonators with $Q=10^4\ldots10^6$ are the quartz crystals
forming the time basis of modern clocks.

\subsection{Arbitrarily driven harmonic oscillator}
In the previous section, we considered the harmonic oscillator with a
sinusoidal or harmonic inhomogeneity, because of the simple temporal
derivatives. The particular solution for the differential equation (\ref
{eq:drivenosc4}) with an inhomogeneity $Ae^{i\omega t}$ was 
\begin{equation}
x_p(t)=\chi(\omega)Ae^{i\omega t}\,,
\end{equation}
with a complex susceptibility $\chi(\omega)$ in
(\ref{eq:complexsusceptibility}). We now consider the more general equation of
motion
\begin{equation}\label{eq:arbitrarydriving}
\ddot{x}+2\beta\dot{x}+\omega_0^2x=a(t)
\end{equation}
and start with an inhomogeneity that is a superposition of two harmonic terms, 
\begin{equation}
a(t)=a_1e^{i\,\Omega_1t}+a_2e^{i\,\Omega_2t}\,.
\end{equation}
Note that the frequencies $\Omega_{1,2}$ can be arbitrary. Due to the linearity of the differential equation, the solution is the
superposition of the solutions for the individual harmonic driving terms,
\begin{equation}
x_p(t)=\chi(\Omega_1)a_1e^{i\,\Omega_1 t}+\chi(\Omega_2)a_2e^{i\,\Omega_2 t}\,.
\end{equation}
This also holds for a more general superposition and its
corresponding solution,
\begin{equation}
a(t)=\sum\limits_{n=-\infty}^\infty a_ne^{i\,\Omega_nt}\quad\longrightarrow\quad
x_p(t)=\sum\limits_{n=-\infty}^\infty a_n\chi(\Omega_n)e^{i\,\Omega_nt}\,.
\end{equation}
For $\Omega_n=n\,\Omega_0$, the sum in $a(t)$ is referred to
as a Fourier series\footnote{after
  \href{https://en.wikipedia.org/wiki/Joseph_Fourier}{Joseph Fourier},
  1768-1830}.
One can show that every square-integrable periodic
function $a(t)$ on an interval $[0,2\pi/\Omega_0[$ can be expressed in this
way\footnote{The right side of this interval is
  open to avoid problems with $\delta$-functions on one of the interval
  limits - Fourier transformations work also for these rather strange functions.}, and the coefficients $a_n$ are uniquely determined by
\begin{equation}
a_n=\frac{\Omega_0}{2\pi}\int\limits_0^{2\pi/\Omega_0}\!\!a(t)\,e^{-i\,n\,\Omega_0t}\,dt\,.
\end{equation}
For non-periodic functions $a(t)$ a similar transformation exists:
\begin{equation}
a(t)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}\!\!\tilde{a}(\omega)\,e^{i\omega
  t}\,d\omega
=:{\cal F}^{-1}\left[\tilde{a}(\omega)\right]\,,
\end{equation}
which is the definition of the inverse Fourier transformation ${\cal F}^{-1}$
for 
the continuous frequency distribution $\tilde{a}(\omega)$. This frequency
distribution can be obtained via the direct Fourier transformation ${\cal F}$,
\begin{equation}
\tilde{a}(\omega)={\cal
  F}\left[a(t)\right]=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}\!\!a(t)\,e^{-i\omega
  t}\,dt\,.
\end{equation}
The particular solution for an arbitrary inhomogeneity $a(t)$ is therefore
given by
\begin{equation}\label{eq:arbitrarysolution}
x_p(t)={\cal F}^{-1}\left[\tilde{x}(\omega)\right]
={\cal F}^{-1}\left[\chi(\omega)\tilde{a}(\omega)\right]
={\cal F}^{-1}\left[\chi(\omega){\cal F}\left[a(t)\right]\right]\,.
\end{equation}
This is a rather simple procedure: First, the Fourier transformation of
the inhomogeneity $a(t)$ is calculated, resulting in a Fourier distribution
$\tilde{a}(\omega)$. The result is multiplied with the complex susceptibility
$\chi(\omega)$, and the product transformed back to obtain $x_p(t)$.

This approach also takes care of the initial conditions
$x(t\rightarrow-\infty)=0$ and  $\dot{x}(t\rightarrow-\infty)=0$, so no
complementary solution needs to be added.

\clearpage
\subsubsection{Examples of Fourier transformations}
A commonly encountered function is a rectangular step function, with a height
$h$ and a width $b$, defined by
\begin{equation}
a(t)=\left\{\begin{array}{rl}h&{\rm for}\,\,|t|<b/2,\\0&{\rm elsewhere.}\end{array}\right.
\end{equation}
The Fourier transform of $a(t)$ can directly be calculated:
\begin{eqnarray}
{\cal F}[a(t)]=\tilde{a}(\omega)&=&
\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty\!\! a(t)e^{-i\omega t}\,dt
=\frac{h}{\sqrt{2\pi}}\int\limits_{-b/2}^{b/2}\!\!e^{-i\omega t}\,dt\nonumber\\
&=&\frac{h}{\sqrt{2\pi}}\,\frac{1}{-i\omega}\,(e^{-i\omega b/2}-e^{+i\omega b/2})=
\frac{h}{\sqrt{2\pi}}\,\frac{2\sin(\omega b/2)}{\omega}\nonumber\\
&=&\frac{hb}{\sqrt{2\pi}}\,\frac{\sin(\omega
  b/2)}{\omega b/2}\,=\,\frac{hb}{\sqrt{2\pi}}\,{\rm sinc}(\omega b/2)\,.
\end{eqnarray}
The characteristic width of the real-valued function $\tilde{a}(\omega)$ in
frequency space is inversely proportional to the width in real space:

\begin{figure*}[h]
\centerline{\includegraphics{sinc.eps}}
\end{figure*}

Another example is the cosine function,
\begin{equation}
a(t)=\cos(\omega_0t)=\frac{1}{2}\left(e^{i\omega_0t}+e^{-i\omega_0t}\right)\,,
\end{equation}
with the rather simple Fourier transform
\begin{equation}
\tilde{a}(\omega)=\sqrt{\frac{\pi}{2}}\left(\delta(\omega-\omega_0)+\delta(\omega+\omega_0)\right)\,,
\end{equation}
which specializes for $\omega_0=0$ into
\begin{equation}
a(t)=1\quad\leftrightarrow\quad\tilde{a}(\omega)=\sqrt{2\pi}\,\delta(\omega)\,,
\end{equation}
with the Dirac delta function $\delta(\omega)$.

\subsubsection{Expression of the oscillator response with Green's function}
While the solution of $x_p(t)$ for an arbitrary inhomogeneity $a(t)$ is
provided by (\ref{eq:arbitrarysolution}), it is not very efficient
in practice, because two integrations have to be carried out: one for the
Fourier transformation, and one for the back transformation. The expression
can be simplified by explicitly writing down both transformations, and then swapping integrations:
\begin{eqnarray}\label{eq:greendef}
x_p(t)&=&{\cal F}^{-1}\Big[ \chi(\omega){\cal
    F}\left[a(t)\right]\Big]\nonumber\\
&=&\frac{1}{2\pi}\int\limits_{-\infty}^\infty\!\!d\omega\,e^{i\omega
  t}\chi(\omega)\int\limits_{-\infty}^\infty\!\!dt'\,e^{-i\omega
  t'}a(t')\nonumber\\
&=&\int\limits_{-\infty}^\infty\!\!dt'\,a(t')\left[\frac{1}{2\pi}\int\limits_{-\infty}^\infty\!\!d\omega\,\chi(\omega)e^{i\omega(t-t')}\right]\nonumber\\
&=&\int\limits_{-\infty}^\infty\!\!dt'\,a(t')g(t-t')\quad{\rm with}\quad
g(\tau):=\frac{1}{2\pi}\int\limits_{-\infty}^\infty\!\!d\omega\,\chi(\omega)e^{i\omega\tau}\,.
\end{eqnarray}
For a given physical system like the damped harmonic oscillator, the function
$g(\tau)$ needs to be evaluated only once, and the result can be used to
obtain $x_p(t)$ from $a(t)$ by a single integration. The combination of $a(t)$
and $g(t)$ in the form above is also referred to as a {\em convolution } of
the two functions $a(t)$ and $g(t)$, which  is
also referred to as \href{https://en.wikipedia.org/wiki/Green's_function}{{\em
    Green's function}}\footnote{after
  \href{https://en.wikipedia.org/wiki/George_Green_(mathematician)}{George
    Green}, 1793-1841} for the physical system.

Green's function has a simple physical interpretation. One rewrites the
inhomogeneity $a(t)$, and compares it with the response $x_p(t)$:
\begin{equation}\label{eq:greenconvo}
a(t)=\int\limits_{-\infty}^\infty\!\!a(t')\delta(t-t')\,dt'\quad\rightarrow\quad
x_p(t)=\int\limits_{-\infty}^\infty\!\!a(t')g(t-t')\,dt'
\end{equation}
Using again the linearity of the differential equation
(\ref{eq:arbitrarydriving}), 
the function $g(t-t')$ presents the response of the harmonic oscillator
to a driving function $\delta(t-t')$, i.e., a delta pulse at time $t'$. An
arbitrary function $a(t)$ can be composed as a superposition of delta 
pulses according to the left side of (\ref{eq:greenconvo}). The response
$x_p(t)$ of the system is an appropriately weighted superposition of the
responses to these delta pulses. This idea behind Green's function carries
over to many other areas in physics, especially in electromagnetism.

The task of determining $g(t)$ for the harmonic oscillator can be solved
in different ways:
\begin{itemize}
\item[(a)] directly by solving the differential equation
  (\ref{eq:arbitrarydriving}) for a delta-shaped driving term, or
\item[(b)] by carrying out the Fourier transformation of the complex
  susceptibility $\chi(\omega)$ according to the definition of $g$ in
  (\ref{eq:greendef}). 
\end{itemize}

\subsubsection{Green's function by  direct integration}
Green's function is the solution of the inhomogeneous differential equation
\begin{equation}\label{eq:deltaresponse}
\ddot{x}\,+\,2\beta\dot{x}\,+\,\omega_0^2x=\delta(t)\,,
\end{equation}
with initial conditions $x(-\epsilon)=0$ and $\dot{x}(-\epsilon)=v(-\epsilon)=0$ for a
small $\epsilon>0$ before the delta-shaped inhomogeneity. During the short
impact at $t=0$, the left side of (\ref{eq:deltaresponse}) can be approximated
by $\ddot{x}(t)$, because the system has not enough time to build up a
significant speed or displacement. Therefore (\ref{eq:deltaresponse}) becomes
\begin{equation}
\ddot{x}=\dot{v}\approx\delta(t)\,.
\end{equation}
This can be directly integrated over a small region around the delta function,
\begin{equation}
\int\limits_{-\epsilon}^{+\epsilon}\dot{v}(t)\,dt=\int\limits_{-\epsilon}^{+\epsilon}\delta(t)\,dt=1\quad\rightarrow\quad
v(\epsilon)-\underbrace{v(-\epsilon)}_{=0}=1\quad{\rm or}\quad v(\epsilon)=1\,.
\end{equation}
The solution $x(t)$ for $t>\epsilon$ can therefore be found by solving the
homogenous differential equation with initial conditions $x(0)=0$ and
$\dot{x}(0)=1$. We have done this already in (\ref{eq:dampedshosolution}).
 Assuming we have a case $\beta<\omega_0$, we choose the form
\begin{equation}
x(t)=\left[B\cos(\omega_1t)+B'\sin(\omega_1 t)\right]e^{-\beta t}\,.
\end{equation}
The initial condition $x(0)=0$ implies that $B=0$. To meet the second initial
condition, we calculate the speed
\begin{equation}
\dot{x}=B'\left[\cos(\omega_1 t)\omega_1e^{-\beta t}+\sin(\omega_1
    t)(-\beta)e^{-\beta t}\right]\quad{\rm or}\quad\dot{x}(0)=B'\omega_1
\end{equation}
and find $B'=1/\omega_1$, which completes
the solution $x(t)$. For $t<0$, we demand $x(t)=0$. Therefore, Green's
function for the damped harmonic oscillator is
\begin{equation}\label{eq:greenresult1}
g(t)=\left\{\begin{array}{rl}\frac{1}{\omega_1}e^{-\beta t}\sin(\omega_1
      t)&{\rm for}\,\,t>0\,,\\0&{\rm for}\,\,t\le0\,.\end{array}\right.
\end{equation}

\begin{figure*}[h]
\centerline{\includegraphics{greenfunc.eps}}
\end{figure*}

\subsubsection{Green's function from Fourier transformation of $\chi(\omega)$}
From the definition in (\ref{eq:greendef}), it follows that Green's function
is essentially the (inverse) Fourier transform of the complex susceptibility:
\begin{equation}\label{eq:fouriersuscept}
g(\tau)\,=\,\frac{1}{\sqrt{2\pi}}{\cal F}^{-1}\left[\chi(\omega)\right]
\,=\,\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\!\!\chi(\omega)\,e^{i\omega\tau}\,d\omega\,=:\,\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}f(\omega)\,d\omega
\end{equation}
Such integrations can be carried out efficiently by making use of a result
from complex calculus. Cauchy's
\href{https://en.wikipedia.org/wiki/Residue_theorem}{\em residue
  theorem}\footnote{after
  \href{https://en.wikipedia.org/wiki/Augustin-Louis_Cauchy}{Augustin-Louis
    Cauchy}, French mathematician, 1789-1857 } considers the integral of a
function $f(z)$ along a closed path $C$ in the complex plane:

\begin{figure*}[h]
\centerline{\includegraphics{complexint2.eps}}
\end{figure*}

The theorem states that the integral of $f(z)$ along $C$ (when evaluated in
counterclockwise direction) is given by
\begin{equation}
\oint_Cf(z)\,dz=2\pi i\sum\limits_{z_k}{\rm Res}(f,z_k)\,,
\end{equation}
where $z_k$ are the {\em poles} of the function $f(z)$ (i.e., locations where
$f(z)$ diverges) inside the path $C$, and ${\rm Res}(f,z_k)$ the so-called
\href{https://en.wikipedia.org/wiki/Residue_(complex_analysis)}{{\em
    residues}} of $f$ 
at the poles $z_k$. Poles outside the path $C$ do not contribute to the
integral. A residue at $z_k$ is defined as element $f_{-1}$ in the Laurent
expansion  
\begin{equation}
f(z)=\sum\limits_{n=-\infty}^\infty f_n\,(z-z_k)^n
\end{equation}
around the position $z_k$ of a pole. This \href{https://en.wikipedia.org/wiki/Laurent_series}{{\em Laurent
series}}\footnote{after
\href{https://en.wikipedia.org/wiki/Pierre_Alphonse_Laurent}{P.A. Laurent},
French mathematician, 1813-1854} is an extension of the
\href{https://en.wikipedia.org/wiki/Taylor_series}{{\em Taylor series}} (where
$n=0\ldots\infty$) to
functions with a singularity/pole.

To use this theorem, we first extend the kernel of the integral
(\ref{eq:fouriersuscept}) to a complex-valued parameter $z$:
\begin{equation}\label{eq:complexkernel}
f(\omega)=\chi(\omega)\,e^{i\omega\tau}\quad\rightarrow\quad
f(z)=\frac{e^{iz\tau}}{\omega_0^2-z^2+2iz\beta}
\end{equation}
The denominator of $f(z)$ vanishes at two complex locations
\begin{equation}
z_{a,b}=i\beta\pm\sqrt{\omega_0^2-\beta^2}=i\beta\pm\omega_1\,,
\end{equation}
where $f(z)$ will have poles:
\begin{equation}
f(z)=\frac{-\,e^{iz\tau}}{(z-z_a)(z-z_b)}\,.
\end{equation}
The minus sign reflects that the $z^2$
term in the denominator of (\ref{eq:complexkernel}) is negative. The two
residues of $f$ are the respective parts that remain when leaving out the
divergent factor $1/(z-z_a)$ or $1/(z-z_b)$:
\begin{eqnarray}
{\rm Res}[f,z_a]&=&\frac{-\,e^{iz_a\tau}}{(z_a-z_b)}
=\frac{-e^{-\beta\tau}\,e^{i\omega_1\tau}}{2\omega_1}\nonumber\\
{\rm Res}[f,z_b]&=&\frac{-\,e^{iz_b\tau}}{(z_b-z_a)}
=\frac{e^{-\beta\tau}\,e^{-i\omega_1\tau}}{2\omega_1}
\end{eqnarray}
The location of the poles of $f$ is shown in the diagram below, together with
the integration path from $z=-\infty$ to $z=+\infty$ along the real axis for (\ref{eq:fouriersuscept}):

\begin{figure*}[h]
\centerline{\includegraphics{complexint1.eps}}
\end{figure*}

In a next step, the integration along the real axis needs to be translated
into an integration along a closed path $C$. For this, we construct
$C$ from a part $C_1$ along the real axis from $z=-R$ to $z=+R$,
and a semicircle $C_2$ with radius $R$:

\begin{figure*}[h]
\centerline{\includegraphics{complexint3.eps}}
\end{figure*}
 
For $R\rightarrow\infty$, the integral along $C_1$ will become the
integral in (\ref{eq:fouriersuscept}) for $g(\tau)$, and we show
that integral $I_2$ along the semicircle $C_2$ vanishes. To do so, we
parameterize  $C_2$ by an angle $\phi$ and the radius $R$: 
\begin{equation}
z=Re^{i\phi}\quad\rightarrow\quad \
I_2=\int_{C_2}f(z)\,dz=\int_{C_2} f(Re^{i\phi})\frac{dz}{d\phi}d\phi\,.
\end{equation}
To show that $I_2$ vanishes, it is enough to consider
the modulus of the integral:
\begin{eqnarray}
|I_2|&=&\left|\int_{C_2}\frac{-e^{iR(\cos\phi+i\sin\phi)\tau}}{(z-z_a)(z-z_b)}iRe^{i\phi}\,d\phi\right|\nonumber\\
&\le&\int_{C_2}\left|\frac{-e^{iR(\cos\phi+i\sin\phi)\tau}}{(z-z_a)(z-z_b)}iRe^{i\phi}\right|\,d\phi=\int_{C_2}\frac{R\,e^{-R\tau\sin\phi}}{\left|(z-z_a)(z-z_b)\right|}\,d\phi
\end{eqnarray}
For $z$ on the semicircle with a radius $R$ large enough to include the poles,
%Todo: add picture to explain why!!!
\begin{equation}
|z-z_{a,b}|> R-\sqrt{\omega_1^2+\beta^2}=R-\omega_0\,,
\end{equation}
and therefore
\begin{equation}
|I_2|\,\le\,\int_{C_2}\frac{R\,e^{-R\tau\sin\phi}}{\left|(z-z_a)(z-z_b)\right|}\,d\phi
\,<\,\frac{R}{(R-\omega_0)^2}\int_{C_2}e^{-R\tau\sin\phi}\,d\phi\,.
\end{equation}

For
$\tau<0$, we choose the bottom semicircle shown in the left part of the
figure. There, $\sin\phi\le0$, so $e^{-R\tau\sin\phi}\le1$ so 
\begin{equation}
|I_2|\,<\,\frac{R}{(R-\omega_0)^2}\int_{C_2}e^{-R\tau\sin\phi}\,d\phi\le\frac{R}{(R-\omega_0)^2}\int_{C_2}\!\!d\phi=\frac{R\,\pi}{(R-\omega_0)^2}\,.
\end{equation}
The last upper bound for $I_2$ vanishes for $R\rightarrow\infty$, so
\begin{equation}
\lim_{R\rightarrow\infty}|I_2|<\lim_{R\rightarrow\infty}\frac{R\,\pi}{(R-\omega_0)^2}=0 
\quad\Rightarrow\quad\lim_{R\rightarrow\infty}I_2=0\,.
\end{equation}
There are no poles surrounded by $C$ in the lower imaginary plane, so
\begin{equation}
g(\tau<0)=\frac{1}{2\pi}\int\limits_{-\infty}^\infty \!\!f(\omega)\,d\omega=
\frac{1}{2\pi}\lim_{R\rightarrow\infty}\int_{C_1}\!\!f(z)\,dz=
\frac{1}{2\pi}\lim_{R\rightarrow\infty}\oint_C\!\!f(z)\,dz=0\,.
\end{equation}

For $\tau>0$, the same argument holds for the vanishing contribution
from $C_2$ in the {\em upper semiplane}. The  
full path $C$ now encloses the two poles at $z_a$ and $z_b$, so
\begin{eqnarray}
g(\tau>0)=\frac{1}{2\pi}\!\int\limits_{-\infty}^\infty\!\!f(\omega)\,d\omega &=&
\frac{1}{2\pi}\lim_{R\rightarrow\infty}\int_{C_1}f(z)dz%\nonumber\\
%&=&
%=\lim_{R\rightarrow\infty}\int_{C_1}f(z)dz+\lim_{R\rightarrow\infty}\int_{C_2}f(z)dz\nonumber\\
%&=&
=\frac{1}{2\pi}\lim_{R\rightarrow\infty}\oint_{C}f(z)\,dz\nonumber\\
&=&\frac{1}{2\pi} 2\pi i \left({\rm Res}[f,z_a]+{\rm Res}[f,z_b]\right)\nonumber\\
&=& i\left(\frac{-e^{-\beta\tau}e^{i\omega_1\tau}}{2\omega_1}+\frac{e^{-\beta\tau}e^{-i\omega_1\tau}}{2\omega_1}\right)\nonumber\\
%&=&2\pi i\frac{e^{-\beta\tau}}{2\omega_i}(-2i\sin\omega_1\tau)\nonumber\\
&=&\frac{1}{\omega_1}e^{-\beta\tau}\sin\omega_1\tau\,.
\end{eqnarray}
This reproduces the result we found by direct integration in
(\ref{eq:greenresult1}).

\subsection{Small oscillations}\label{sec:smalloscillations}
The concept of the harmonic oscillator and its associated solutions in the
presence of damping and external excitation can be found in many
physical situations, and may apply even where a restoring force is only
approximately linear in the respective coordinate. We
consider an arbitrary potential for a system with a single degree of freedom,
describing the conservative interaction with its environment:

\begin{figure*}[h]
\centerline{\includegraphics{potential2.eps}}
\end{figure*}

The potential $U(x)$ shall have a minimum at position $x_1$. For small
deviations $u$ from $x_1$, the potential can be approximated by a
Taylor expansion:
\begin{equation}
U(x)=U(x_1)\,+\,\underbrace{\left.\frac{\partial U}{\partial
    x}\right|_{x=x_1}\!\!\!\!\!\!u}_{=0}\,+\,\frac{1}{2}\left.\frac{\partial^2
    U}{\partial x^2}\right|_{x=x_1}\!\!\!\!\!\!u^2\,+\,\ldots\,,\quad {\rm
  with}\,\, u=(x-x_1)
\end{equation}
In a minimum of $U(x)$, the second term vanishes by definition, and the
potential resembles that of a harmonic oscillator, since the constant offset
$U(x_1)$ does not change the dynamics of the system.

The minimum is characterized by a positive second derivative of the potential,
\begin{equation}
\left.\frac{\partial^2 U}{\partial x^2}\right|_{x=x_1}>0\,,
\end{equation}
leading to the oscillatory solutions of the harmonic oscillator.

In presence of a damping force, a system starting near $x_1$ with a low enough
velocity will approach $x_1$ asymptotically in time. Therefore, such a
position is referred to as a {\em stable equilibrium}. For a maximum in the
potential energy, like at position $x_2$ in the graph above, the
second derivative is negative, which does not lead to an oscillatory
motion. While a particle at rest at $x_2$ will stay there for all times, any
small deviation will lead to a trajectory that starts with an exponentially
growing separation from $x_2$. 

As most of the physical systems will experience small fluctuations of forces,
and some dissipation, system tend to evolve into a state near a minimum where
$\partial^2U/\partial{x}^2>0$, which is therefore also referred to as a {\em
  stable equilibrium}. A position where the potential energy
takes a maximum, or $\partial^2U/\partial{x}^2<0$ is referred to as an
{\em unstable equilibrium}. To complete the description, a location like $x_3$
where $\partial^2U/\partial{x}^2=0$ over some extended interval
is referred to as an {\em indifferent equilibrium}. It should be pointed out
that a minimum or maximum could still 
be present, but the second derivative could vanish. In that case, the sign of
the first non-vanishing term in the Taylor expansion of $U(x)$ would determine
if the equilibrium is stable or unstable, but such situations are very rare in
practice.

Therefore, the small deviations of a system from a stable equilibrium can
often be mapped to a harmonic oscillator, assuming the kinetic energy is of
a quadratic form in the velocity $\dot{x}$. As the restoring force
$F=-\partial U/\partial x$ near a minimum is linear in the deviation $u$, this
procedure is sometimes referred to as {\em  linearization}.

\subsubsection{Example of the plane pendulum}
As a simple example, we revisit the plane pendulum from section
\ref{sec:pendulum}:\vspace{0.5em}

\raisebox{-1.4cm}{\includegraphics{pendulum2.eps}}\hfill
\begin{minipage}{10cm}
The potential and kinetic energy were given by
\begin{eqnarray}
U&=&mgl\,(1-\cos\theta)\,,\nonumber\\
T&=&\frac{1}{2}\,ml^2\,\dot{\theta}^2\,,
\end{eqnarray}
leading to the nonlinear differential equation (\ref{eq:eqmotionpendulum1})
\begin{equation}
\ddot\theta+\frac{g}{l}\sin\theta=0\,.
\end{equation}
\end{minipage}\vspace{0.5em}

\noindent A Taylor expansion of the potential near the minimum at $\theta=0$ leads to
\begin{equation}
U(\theta)\approx\frac{1}{2}\,mgl\,\theta^2\,,
\end{equation}
and an approximate Lagrange function of
\begin{equation}
L=T-U=\frac{1}{2}\,ml^2\,\dot{\theta}^2-\frac{1}{2}\,mgl\,\theta^2\,,
\end{equation}
which after some simplification leads to the equation of motion
\begin{equation}
\ddot\theta\,+\,\frac{g}{l}\,\theta=0\,.
\end{equation}
This is the equation of motion for the undamped harmonic oscillator, and by
comparison  e.g. with (\ref{eq:dampedshoeom}), one can immediately extract the
oscillation frequency
\begin{equation}
\omega_0=\sqrt{\frac{g}{l}}\,.
\end{equation}
\clearpage

\section{Coupled oscillations}
So far, we have encountered methods to generate equations of motion for system
of many particles, but have not really solved very complex systems. In this
section, we will look into a typical system of many harmonic oscillators. Such
a problem may arise from a system of particles near an equilibrium that are
coupled together by some localized interactions, and can be approximated by
harmonic oscillators as seen in section \ref{sec:smalloscillations}.

\subsection{Two coupled oscillators}\label{ssec:twocoupledoscillators}
We demonstrate the strategy to tackle coupled oscillator problems with that of
two masses, coupled by springs:

\begin{figure*}[h]
\centerline{\includegraphics{twomasses.eps}}
\end{figure*}

The equations of motion can be obtained in various ways, and form a system of
{\em coupled} differential equations:
\begin{eqnarray}
m\,\ddot{x}_1+k_1x_1+k_{12}(x_1-x_2)&=&0\nonumber\\
m\,\ddot{x}_2+k_2x_2+k_{12}(x_2-x_1)&=&0\label{eq:twomassesEoM0}
\end{eqnarray}
The coupling means that these are not independent equations of motion for
$x_1$ and $x_2$, and we need to find a solution for both variables that
satisfy both equations at the same time. As these equations are still linear,
we can try the previous trick, and look for complex solutions of the form
\begin{equation}
x_1(t)=b_1e^{i\omega t}\,,\quad x_2(t)=b_2e^{i\omega t}\,,
\end{equation}
with the same oscillation frequency $\omega$ for both variables, but different
complex amplitudes $b_1$, $b_2$. Inserting these into (\ref{eq:twomassesEoM0})
leads to
\begin{eqnarray}
\left(-m\omega^2b_1\,+\,(k_1+k_{12})b_2\,-\,k_{12}b_1\right)e^{i\omega
  t}&=&0\nonumber\\
\left(-m\omega^2b_2\,+\,(k_2+k_{12})b_1\,-\,k_{12}b_2\right)e^{i\omega
  t}&=&0\,.
\end{eqnarray}
As before, we can divide by the exponential function, and obtain an algebraic
equation determining the oscillation frequency. This time,
however, the two equations remain coupled via the amplitudes. The
algebraic equation is linear in the $b$, and can be written in matrix form,
\begin{equation}
\left(\begin{array}{cc}\label{eq:twomassesEoM}
-m\omega^2+(k_1+k_{12})&-k_{12}\\-k_{12}&-m\omega^2+(k_2+k_{12})
\end{array}\right)\cdot\left(\begin{array}{c}b_1\\b_2\end{array}\right)=0\,,
\end{equation}
or simply
\begin{equation}
\tensor{M}\cdot\vec{b}=0\,,
\end{equation}
where the symbol $\tensor{M}$ denotes a matrix, and $\vec{b}$ a vector that is
made up by two amplitudes. Note that this is now a vector that does not
represent a vector in the usual three-dimensional space, it just stores the
coefficients. The condition for this matrix equation to be fulfilled is 
\begin{equation}
{\rm det}\,\tensor{M}=0\,,
\end{equation}
which leads to the characteristic equation of the linear equation
system (\ref{eq:twomassesEoM})
\begin{equation}
\left[-m\omega^2+(k_1+k_{12})\right]\left[-m\omega^2+(k_2+k_{12})\right]-k_{12}^2=0\,.
\end{equation}
To simplify the subsequent treatment, we assume now that $k_1=k_2=k$, so the
characteristic equation becomes
\begin{equation}
(k+k_{12}-m\omega^2)^2-k_{12}^2=0\,,
\end{equation}
which can easily be resolved for the two roots for $\omega$,
\begin{equation}\label{eq:twofrequencies}
\omega_{1,2}=\sqrt{\frac{k+k_{12}\pm k_{12}}{m}}\qquad{\rm or}\quad
\omega_1=\sqrt{\frac{k+2k_{12}}{m}}\,,\quad\omega_2=\sqrt{\frac{k}{m}}\,.
\end{equation}
As with the simple harmonic oscillator, we can have solutions with both signs
for $\omega$, and the general solution of the system (\ref{eq:twomassesEoM})
would be given by
\begin{eqnarray}\label{eq:twomassessol1}
x_1(t)&=&b_{11}^+e^{i\omega_1t}+b_{11}^-e^{-i\omega_1t}+b_{12}^+e^{i\omega_2t}+b_{12}^-e^{-i\omega_2t}\nonumber\\
x_2(t)&=&b_{21}^+e^{i\omega_1t}+b_{21}^-e^{-i\omega_1t}+b_{22}^+e^{i\omega_2t}+b_{22}^-e^{-i\omega_2t}\,.
\end{eqnarray}
Here, the indices on $b$ indicate the mass index, and the frequency of the
system. However, these eight coefficients are not independent - they still
need to fulfill (\ref{eq:twomassesEoM}), which imposes a relation between the
amplitudes for the two masses. Inserting the solutions
(\ref{eq:twomassessol1}), and restricting to either positive or negative
frequencies yields
\begin{eqnarray}
{\rm
  for}\,\,\omega=\omega_1:&\quad&-k_{12}b_{11}-k_{12}b_{21}=0\quad\rightarrow\quad
b_{11}=-b_{21}=:B_1\,,\nonumber\\
{\rm
  for}\,\,\omega=\omega_2:&\quad&+k_{12}b_{12}-k_{12}b_{22}=0\quad\rightarrow\quad
b_{12}=b_{22}=:B_2\,.
\end{eqnarray}
The general solution of the coupled problem is then given by
\begin{eqnarray}\label{eq:solutiontwomasses}
x_1(t)&=&B_1^+e^{i\omega_1t}+B_1^-e^{-i\omega_1t}+B_2^+e^{i\omega_2t}+B_2^-e^{-i\omega_2t}\nonumber\\
x_2(t)&=&-B_1^+e^{i\omega_1t}-B_1^-e^{-i\omega_1t}+B_2^+e^{i\omega_2t}+B_2^-e^{-i\omega_2t}\,,
\end{eqnarray}
with four constants $B_1^+$, $B_1^-$, $B_2^+$, $B_2^-$ to satisfy the initial
conditions of the problem --- the system is hereby completely determined, as
(\ref{eq:twomassesEoM}) is a system of differential equations of second order.

\subsubsection{Normal coordinates}\label{ssec:normalcoordinates1}
The system has two oscillation frequencies $\omega_1$ and $\omega_2$, and both
masses participate in oscillations at those two frequencies. One can ask
for coordinates that simplify the description of the
problem. To see that this is the case, we define coordinates
\begin{equation}\label{eq:normalcoord1}
\eta_1:=x_1-x_2\,,\quad{\rm and}\quad\eta_2:=x_1+x_2\,,
\end{equation}
such that the original coordinates can be obtained again via
\begin{equation}
x_{1,2}=\frac{1}{2}(\eta_2\pm\eta_1)\,.
\end{equation}
Inserting this into the coupled equations of motion (\ref{eq:twomassesEoM0})
leads to
\begin{eqnarray}
\frac{m}{2}(\ddot{\eta}_2+\ddot{\eta}_1)\,+\,\frac{k}{2}(\eta_2+\eta_1)\,+\,k_{12}\eta_1&=&0\nonumber\\
\frac{m}{2}(\ddot{\eta}_2-\ddot{\eta}_1)\,+\,\frac{k}{2}(\eta_2-\eta_1)\,-\,k_{12}\eta_1&=&0
\end{eqnarray}
This system of equations can be transformed into another one by
subtracting and adding the two equations,
\begin{eqnarray}
m\,\ddot{\eta}_1\,+\,(k+2k_{12})\,\eta_1&=&0\nonumber\\
m\,\ddot{\eta}_2\,+\,k\,\eta_2&=&0\,.
\end{eqnarray}
This is now a {\em decoupled} system of two equations of two independent
harmonics oscillators. Their solutions can be simply written from
(\ref{eq:lineq:sol2}) as
\begin{eqnarray}
\eta_1(t)&=&C_1^+e^{i\omega_1t}\,+\,C_1^-e^{-i\omega_1t}\,,\nonumber\\
\eta_2(t)&=&C_2^+e^{i\omega_2t}\,+\,C_2^-e^{-i\omega_2t}\,.
\end{eqnarray}
Coordinates $\eta_1$ and $\eta_2$ for this system are referred to as {\em
  normal coordinates}, a name that is justified in a later section. By
appropriate choice of initial conditions, one can ensure that only one of the
coordinates has a non-trivial solution. For example, if we choose
\begin{equation}
x_1(0)=-x_2(0)\quad{\rm and}\quad \dot{x}_1(0)=-\dot{x}_2(0)\,,
\end{equation}
we know from (\ref{eq:normalcoord1}) that 
\begin{equation}
\eta_2(0)=0\quad{\rm and}\quad \dot{\eta}_2(0)=0\,,
\end{equation}
and we will find an oscillation with a single frequency $\omega_1$. For this
solution, the two amplitudes $x_1$ and $x_2$ are always related via
\begin{equation}
x_1(t)=-x_2(t)\,,
\end{equation}
i.e., the two masses move {\em out of phase} or in an antisymmetric
oscillation; occasionally this mode of oscillation is also referred to as a
{\em breathing mode}. Similarly, if initial conditions are chosen such that
\begin{equation}
x_1(0)=x_2(0)\quad{\rm and}\quad \dot{x}_1(0)=\dot{x}_2(0)\,,
\end{equation}
an oscillation takes place only at a frequency $\omega_2$, with
\begin{equation}
x_1(t)=x_2(t)
\end{equation}
at all times. This mode of oscillation is a symmetric mode, and some times
referred to as {\em common mode oscillation}.

Both oscillation modes with only one frequency appearing can be understood as
an effective one-variable problem: For the antisymmetric mode, the individual
masses oscillate independently between two springs and a fixed 
center position of the middle spring, leading to a larger effective spring
constant for the single mass motion. For the symmetric case, the inner spring
stays always in its equilibrium position, and the only restoring force to the
masses motion is provided by the outside springs, leading to the same result
as we have seen from the simple mass/spring system in section
\ref{sssec:harmonicosc1}.

\subsubsection{Beat of oscillations}
Before we move on to a more general treatment of coupled systems, we consider
the special case where the coupling constant $k_{12}$ between the two masses
is much smaller than the other spring constants $k$. Then, the two oscillation
frequencies $\omega_{1,2}$ from (\ref{eq:twofrequencies}) become very similar,
and one can define
\begin{equation}
\omega_0:=\frac{\omega_1+\omega_2}{2}\,,\quad{\rm and}\quad\Delta:=\frac{\omega_1-\omega_2}{2}\,,
\end{equation}
with $\omega_0\gg\Delta$. If we further assume the initial conditions
\begin{equation}
x_1(0)=D\,,\quad x_2(0)=0\,,\quad \dot{x}_1(0)=\dot{x}_2(0)=0\,,
\end{equation}
i.e., the system starts with only one mass displaced from its resting
position, the coefficients in (\ref{eq:solutiontwomasses}) become
\begin{equation}
B_1^+=B_1^-=B_2^+=B_2^-=\frac{D}{4}\,.
\end{equation}
The oscillation of mass 1 over time can be written as
\begin{eqnarray}
x_1(t)&=&\frac{D}{4}\left[e^{i\omega_1t}+e^{-i\omega_1t}+e^{i\omega_2t}+e^{-i\omega_2t}\right]\nonumber\\
&=&\frac{D}{4}\left(2\cos(\omega_1t)+2\cos(\omega_2t)\right)\nonumber\\
&=&D\cos\left[\frac{\omega_1+\omega_2}{2}\right]\,
\cos\left(\frac{\omega_1-\omega_2}{2}\right)\,=\,D\cos(\omega_0 t)\cos(\Delta t)\,.
\end{eqnarray}
Similarly, one obtains for the oscillation of mass 2 the expression
\begin{equation}
x_2(t)=D\sin(\omega_0 t)\sin(\Delta t)\,.
\end{equation}
Qualitatively, the oscillation of the system looks as shown below:

\begin{figure*}[h]
\centerline{\includegraphics{beats.eps}}
\end{figure*}
Initially, mass 1 oscillates with approximate frequency $\omega_0$, and mass 2
is at rest. Then, the oscillation of mass 1 decreases in amplitude, while the
amplitude of mass 2 increases. At time $t=\pi/2\Delta$, only mass 2 oscillates,
before the pattern reverses, and mass 1 increases its amplitude again. This
phenomenon is referred to as a {\em beat} of two oscillations, and can be
found in many weakly coupled oscillators of the same frequency. The dashed
lines in the figure above indicate the envelope of the oscillations.

\subsection{Many coupled particles -- small oscillations}
In this section, we try to extend the idea of small oscillations to a system
of many particles, e.g. molecules or the even
more complex solids, where many atoms are hold together by chemical bonds,
which are not completely rigid. However, the approach we take is not limited
to these special situations, and goes also beyond mechanics.

We start with a system described by $N$ generalized
coordinates, forming the set $\{q_k\}$. The interactions should be conservative,
i.e., they can be described by a single potential $U(\{q_k\})$. We now assume
that the system is near an equilibrium state, possibly due to the presence of
some dissipating forces which we exclude in the treatment. The
equilibrium position of all coordinates shall be given by the set
$\{q_{k,0}\}$. Similar to the simple one-dimensional problem in section
\ref{sec:smalloscillations}, we can approximate the potential $U$ by a
truncated Taylor expansion. This time, however, we need to do the Taylor
expansion for $N$ coordinates:
\begin{equation}
  U(\{q_k\})=\underbrace{U(\{q_{k,0}\})}_{=:U_0}\,+\,\sum\limits_k\underbrace{\left.\frac{\partial U}{\partial
      q_k}\right|_0}_{=-\phi_k}\!\!(q_k-q_{k,0})\,+\,\sum\limits_{j,k}\left.\frac{\partial^2
    U}{\partial q_j\partial q_k}\right|_0
\frac{1}{2}(q_k-q_{k,0})(q_j-q_{j,0})+\ldots
\end{equation}
The first term is simply the constant potential energy $U_0$ in the
equilibrium, and will not be relevant for the dynamics of the system. In the
second term, the first derivative of of the potential corresponds to the
generalized force $\phi_k$ (see section
\ref{ssec:nonconservativeforces}). These forces are evaluated at the
equilibrium position $\{q_{k,0}\}$ of the system, where they vanish by
definition. By introducing variables
\begin{equation}
u_k:=q_k-q_{k,0}
\end{equation}
for the displacement of coordinate $k$ from the equilibrium position, we can
formulate the approximate potential energy by the first non-vanishing and
relevant term in these displacements:
\begin{equation}\label{eq:elasticitymatrix}
U=U_0+\frac{1}{2}\sum\limits_{j,k}A_{jk}u_ju_k\,,\quad{\rm with}\quad
A_{jk}:=\left.\frac{\partial^2U}{\partial q_j\partial q_k}\right|_{0}\,,
\end{equation}
where the index 0 at the second derivative in $A_{jk}$ indicates again that it
has to be taken at the equilibrium position set $\{q_{k,0}\}$. Since the
sequence of the differentiation does not matter, we have the symmetry
\begin{equation}
A_{jk}=A_{kj}\,.
\end{equation}

We now try a similar expansion for the kinetic energy $T$ of the system. Here,
we recall from (\ref{eq:kineticenergy2}) in section
\ref{ssec:HamiltonTotalenergy} that for time-independent transformations from
Cartesian to generalized coordinates, the total kinetic energy can be written as
\begin{equation}\label{eq:coupledT}
T=\sum\limits_{j,k}a_{jk}\dot{q}_j\dot{q}_k\quad{\rm
  with}\quad a_{jk}=\frac{1}{2}\sum\limits_\alpha m_\alpha \sum\limits_i\frac{\partial
  x_{\alpha,i}}{\partial q_k}\frac{\partial x_{\alpha,i}}{\partial q_j}\,,
\end{equation}
where the summations in the definition of $a_{jk}$ go over all particles
$\alpha$ and Cartesian coordinates $i$. Since the $a_{jk}$ still can depend on
the coordinates, we also perform a Taylor expansion,
\begin{equation}
a_{jk}=\left.a_{jk}\right|_0\,+\,\sum\limits_l\left.\frac{\partial a_{jk}}{q_l}\right|_0\!\!(q_l-q_{l,0})\,+\,\ldots
\end{equation}
In this expansion, the first term {\em does not} vanish, so we stop right
there (even neglecting the linear term), and define
\begin{equation}\label{eq:massmatrixelement}
m_{jk}:=2\left.a_{jk}\right|_0\,.
\end{equation}
These coefficients do not depend on the deviations $u_k$ anymore. 
Because the sequence of the differentiations in $a_{jk}$ does not matter, we
also have $m_{jk}=m_{kj}$. With $\dot{q}_k=\dot{u}_k$, the kinetic
energy can then be written as
\begin{equation}
T=\frac{1}{2}\sum\limits_{j,k}m_{jk}\dot{u}_j\dot{u}_k\,.
\end{equation}
The Lagrange function for small deviations from the equilibrium position
becomes
\begin{equation}\label{eq:coupledlagrange1}
L=T-U=\frac{1}{2}\sum\limits_{i,j}\left[m_{ij}\dot{u}_i\dot{u}_j-A_{ij}u_iu_j\right]-U_0\,,
\end{equation}
with {\em constant} coefficients $m_{ij}$ and $A_{ij}$. This expression has
bilinear terms in the new coordinates $u_k$, and bilinear terms in the
velocities $\dot{u}_k$ --- a
structure very similar to a harmonic oscillator. The resulting equations of
motion are obtained via the Euler-Lagrange method:
\begin{equation}
\frac{\partial L}{\partial u_k}-\frac{d}{dt}\frac{\partial L}{\partial\dot{u}_k}
=-\sum\limits_jA_{jk}u_j-\frac{d}{dt}\sum\limits_jm_{jk}\dot{u}_j\,=\,0\,,
\end{equation}
or
\begin{equation}
\sum\limits_j\Big[m_{jk}\ddot{u}_j+A_{jk}u_j\Big]=0\quad{\rm for\,\,all}\,\,k\,.
\end{equation}
This set of equations can be written as a matrix equation,
\begin{equation}\label{eq:coupledEOM}
\tensor{m}\cdot\ddot{\vec{u}}(t)+\tensor{A}\cdot\vec{u}(t)=0\,,
\end{equation}
where $\tensor{m}$ is the {\em mass matrix}, $\tensor{A}$ is a matrix that
describes the elastic response of a system, and $\vec{u}$ is a vector made up
by all the displacement coordinates $u_k$ in the system. 
\subsubsection{Solving the equations of motion}
We can solve
equation (\ref{eq:coupledEOM}) in the same way as in section
\ref{sssec:harmonicosc1}, e.g.\ with a harmonic ansatz
\begin{equation}
\vec{u}(t)=\vec{a}\cos(\omega t-\delta)\,,\quad{\rm or\,\,componentwise:}\quad
u_k(t)=a_k\cos(\omega t-\delta)\,,
\end{equation}
leading to an algebraic set of equations
\begin{eqnarray}\label{eq:eigenfrequencies}
(\tensor{A}-\omega^2\tensor{m})\cdot\vec{a}&=&0\\
{\rm or\,\,componentwise:}\quad
\sum\limits_j(A_{kj}-\omega^2m_{kj})a_j&=&0\,\,\forall_k\,,
\end{eqnarray}
with constant amplitude coefficients $a_k$ forming a vector $\vec{a}$. This is
again a set of algebraic equations, with a condition
\begin{equation}
{\rm det}\,(\tensor{A}-\omega^2\tensor{m})\stackrel{!}{=}0
\end{equation}
to have solutions. This characteristic or secular
equation for the $\omega^2$ has $N$ roots, where $N$ is the number of
coordinates in the system. The resulting frequencies $\omega_r, r=1\ldots N$
are referred to as {\em eigenfrequencies} or characteristic frequencies of the
problem.

For each eigenfrequency $\omega_r$, a vector $\vec{a}_r$ of
amplitude coefficients solves the equation set
(\ref{eq:eigenfrequencies}). These vectors characterize the modes of
oscillation, i.e., the relative amplitude with which each coordinate
oscillates at this particular frequency. That mode of oscillation is also
referred to as {\em eigenmode} for an oscillation at frequency
$\omega_r$. With this, the general solution of the coupled oscillation can be
written as
\begin{equation}\label{eq:coupledsolution}
\vec{u}(t)=\sum\limits_{r=1}^N\alpha_r\vec{a}_r\cos(\omega_rt-\delta_r)\,,
\end{equation}
with a factor $\alpha_r$ permitting normalization of $\vec{a}_r$, or componentwise
\begin{equation}
u_k(t)=\sum\limits_{r=1}^N\alpha_ra_{k,r}\cos(\omega_r t-\delta_r)\,,
\end{equation}
with real-valued amplitudes $\alpha_r\vec{a}_r$ and phase shifts $\delta_r$ for the
contributions of the various eigenmodes to the oscillation. In the
componentwise expression, the elements $a_{k,r}$ are the $k$-th component of
eigenvector $\vec{a}_r$.

It is interesting to investigate some properties of the vector
$\vec{a}_r$ that characterizes a particular eigenmode. Since it has to fulfill
(\ref{eq:eigenfrequencies}), it is not completely determined, and any
vector $\alpha\,\vec{a}_r$ would also fulfill that equation. One can therefore
choose the normalization of the vector such that
\begin{equation}\label{eq:eignevectornormalization}
\vec{a}_r\cdot(\tensor{m}\cdot\vec{a}_r)=1\,.
\end{equation}
In this expression, the term in the parenthesis is a vector multiplied from
the right to a matrix, leading to another vector. This vector then gets
multiplied with another vector via a scalar product, leading to a scalar
result 1. Furthermore, vectors $\vec{a}_r$ and $\vec{a}_s$ fulfill a generalized
orthogonality relation\footnote{Careful: this is not the usual orthogonality
  relation between two vectors, which would be
  $\vec{a}_r\cdot\vec{a}_s=\delta_{rs}$. The two relations differ if the
  masses $m_\alpha$ are not all the same.}:
\begin{equation}\label{eq:orhto0}
\vec{a}_r\cdot\tensor{m}\cdot\vec{a}_s=0\quad{\rm for}\quad r\neq s\,.
\end{equation}
To see this, we take two solutions of (\ref{eq:eigenfrequencies}) for
different eigenfrequencies,
\begin{eqnarray}
\omega_r^2\,\tensor{m}\cdot\vec{a}_r&=&\tensor{A}\cdot\vec{a}_r\nonumber\\
\omega_s^2\,\tensor{m}\cdot\vec{a}_s&=&\tensor{A}\cdot\vec{a}_s\,,
\end{eqnarray}
and multiply these equations from the right with $\vec{a}_s$ and
$\vec{a}_r$, respectively:
\begin{eqnarray}\label{eq:orhto1}
\omega_r^2\,\vec{a}_s\cdot\tensor{m}\cdot\vec{a}_r&=&\vec{a}_s\cdot\tensor{A}\cdot\vec{a}_r\nonumber\\
\omega_s^2\,\vec{a}_r\cdot\tensor{m}\cdot\vec{a}_s&=&\vec{a}_r\cdot\tensor{A}\cdot\vec{a}_s\,.
\end{eqnarray}
Evaluating one of the matrix sandwiched between two vectors in
components, and using the symmetry $A_{jk}=A_{kj}$,
\begin{eqnarray}
\vec{a}_s\cdot\tensor{A}\cdot\vec{a}_r&=&\sum\limits_{j,k}a_{j,s}\,A_{jk}\,a_{k,r}
= \sum\limits_{j,k}a_{j,s}\,A_{kj}\,a_{k,r} \nonumber\\ 
&=&\sum\limits_{j,k}a_{k,r}\,A_{kj}\,a_{j,s}\,=\,\vec{a}_r\cdot\tensor{A}\cdot\vec{a}_s\,,
\end{eqnarray}
we see that the two right sides of
(\ref{eq:orhto1}) are the same. Similarly, due to $m_{kj}=m_{jk}$, the
sandwich products $\vec{a}_s\cdot\tensor{m}\cdot\vec{a}_r$ and
$\vec{a}_r\cdot\tensor{m}\cdot\vec{a}_s$ on the left sides of
(\ref{eq:orhto1}) are the same, so one can subtract the two equations and
obtain
\begin{equation}
(\omega_s^2-\omega_r^2)\,\vec{a}_r\cdot\tensor{m}\cdot\vec{a}_s=0\,.
\end{equation}
Assuming that the eigenfrequencies are not degenerate, the difference of its
squares in the parenthesis does not vanish for $r\neq s$, and therefore, the
sandwich product must vanish. This means that the two vectors obey the
generalized orthogonality relation (\ref{eq:orhto1}), which can be combined
with the normalization (\ref{eq:eignevectornormalization}) to
\begin{equation}\label{eq:orhtonormality1}
\vec{a}_r\cdot\tensor{m}\cdot\vec{a}_s=\delta_{rs}\,,
\end{equation}
with the Kronecker symbol $\delta_{rs}$.

The orthogonality of eigenvectors is one of the results of
linear algebra; in fact, the
whole search for the oscillation modes can be mapped to an eigenvector
problem. To see this, we first recognize that the matrix $\tensor{m}$ can be
inverted. For the simple case that the ${q_k}$ are Cartesian
coordinates, one can see from the definitions (\ref{eq:coupledT}) and
(\ref{eq:massmatrixelement}) of the matrix elements $m_{jk}$ of
$\tensor{m}$ that there are no off-diagonal elements, and the diagonal entries
in $\tensor{m}$ are simply the masses corresponding to coordinate $k$:
\begin{equation}
m_{jk}=\delta_{jk}m_k\,.
\end{equation}
Then, $\tensor{m}$ can be inverted\footnote{In fact, $\tensor{m}$ can always
  be inverted if there are no redundant coordinates.}, with matrix elements of the inverse
matrix $\tensor{m}^{-1}$
\begin{equation}
(\tensor{m}^{-1})_{jk}=\delta_{jk}1/m_k\,.
\end{equation}
Therefore, we can multiply equation (\ref{eq:eigenfrequencies}) from the left
with $\tensor{m}^{-1}$, and obtain
\begin{equation}
(\tensor{m}^{-1}\cdot\tensor{A})\cdot\vec{a}=\omega^2\vec{a}\,,
\end{equation}
which is the familiar eigenvector/eigenvalue equation from linear algebra for
the matrix $\tensor{m}^{-1}\cdot\tensor{A}$, which is a $N\times N$ matrix if
there are $N$ degrees of freedom. There are $N$ eigenvalues $\omega_r^2$, and
the corresponding eigenvectors $\vec{a}_r$ determine the oscillation modes. If
the eigenvalues of a matrix are not degenerate, the corresponding eigenvectors
are orthogonal. If there are degenerate eigenvalues, the subset of the
corresponding eigenvectors is orthogonal to all other eigenvectors, and a
linear combination of eigenvectors in that subspace can be found such that all
eigenvectors are orthogonal. As the orthogonal vectors $\vec{a}_r$
characterize the oscillation modes to an eigenfrequency $\omega_r$, the modes
are characterized by $\vec{a}_r$ are also called {\em normal modes}.

\subsubsection{Normal coordinates}
In section \ref{ssec:normalcoordinates1} we introduced normal coordinates for
the problem of two masses on an ad-hoc basis, and saw that the choice indeed
decoupled the equations of motion. We can do this now for the general
case, and introduce the connection between individual displacements $u_k$ and
normal coordinates $\eta_r$:
\begin{equation}\label{eq:normalcoord2}
u_k(t)=\sum\limits_ra_{k,r}\eta_r(t)\,,
\end{equation}
or in vectorial form
\begin{equation}\label{eq:normalcoord3}
\vec{u}=\sum\limits_r\vec{a}_r\eta_r(t)\,.
\end{equation}
The transformation from the original coordinates $u_k$ to normal coordinates
can be accomplished by realizing that the eigenvectors $\vec{a}_r$ are all
orthogonal; when using normalization (\ref{eq:orhtonormality1}), one can
multiply the relation (\ref{eq:normalcoord2}) from the left with
$\vec{a}_s\cdot\tensor{m}$ to directly obtain the normal coordinate
$\eta_s$:
\begin{eqnarray}
\vec{a}_s\cdot\tensor{m}\cdot\vec{u}&=&\vec{a}_s\cdot\tensor{m}\cdot\sum\limits_r\vec{a}_r\eta_r\nonumber\\
&=&\sum\limits_r\underbrace{\vec{a}_s\cdot\tensor{m}\cdot\vec{a}_r}_{=\delta_{rs}}\eta_r\nonumber\\
&=&\eta_s\,.
\end{eqnarray}
This helps e.g.\ to express an initial condition given in $\vec{u}$ in the
normal coordinates.
Since the coefficient matrices $\tensor{m}$ and $\tensor{A}$ do not depend on
time, the velocities are
\begin{equation}
\dot{u}_k(t)=\sum\limits_ra_{k,r}\dot{\eta}_r(t)\quad{\rm or}\quad
\dot{\vec{u}}=\sum\limits_r\vec{a}_r\dot{\eta}_r(t)\,.
\end{equation}
The coupled Lagrange function in (\ref{eq:coupledlagrange1}) can also be
expressed in matrix form,
\begin{eqnarray}
L&=&\frac{1}{2}\sum\limits_{i,j}\Big[m_{ik}\dot{u}_i\dot{u}_j-A_{ij}u_iu_j\Big]\nonumber\\
&=&\frac{1}{2}\Big[\dot{\vec{u}}\cdot(\tensor{m}\cdot\dot{\vec{u}})-\vec{u}\cdot\tensor{A}\cdot\vec{u}\Big]\,,
\end{eqnarray}
assuming without loss of generality that $U_0=0$. Using expression
(\ref{eq:normalcoord3}) to make the transition to normal coordinates leads to
\begin{equation}
L=\frac{1}{2}\sum\limits_{r,s}\dot{\eta}_r\dot{\eta}_s\underbrace{\vec{a}_r\cdot\tensor{m}\cdot\vec{a}_s}_{=\delta_{rs}}-\frac{1}{2}\sum\limits_{r,s}\eta_r\eta_s\vec{a}_r\cdot\tensor{A}\cdot\vec{s_s}\,,
\end{equation}
where the first sandwich product is just the orthogonality relation
(\ref{eq:orhtonormality1}). For the second term, we use
(\ref{eq:eigenfrequencies}) for eigenvector $\vec{a}_s$,
\begin{equation}
\tensor{A}\cdot\vec{a}_s=\omega_s^2\tensor{m}\cdot\vec{a}_s\,,
\end{equation}
and continue with the evaluation of the Lagrange function:
\begin{eqnarray}
L&=&\frac{1}{2}\sum\limits_r\dot{\eta}_r^2-\frac{1}{2}\sum\limits_{r,s}\eta_r\eta_s\vec{a}_r\omega_s^2\tensor{m}\cdot\vec{a}_s\nonumber\\
&=&\frac{1}{2}\sum\limits_r\dot{\eta}_r^2-\frac{1}{2}\sum\limits_{r,s}\eta_r\eta_s\omega_s^2\underbrace{\vec{a}_r\cdot\tensor{m}\cdot\vec{a}_s}_{=\delta_{rs}}\nonumber\\
&=&\frac{1}{2}\sum\limits_r\Big[\dot{\eta}_r^2-\eta_r^2\omega_r^2\Big]\,.
\end{eqnarray}
This is a sum of Lagrange functions for simple harmonic oscillators, which
means that the motion in normal coordinates is completely decoupled. The
corresponding equations  of motion via Euler-Lagrange are
\begin{equation}
\frac{\partial L}{\partial\eta_r}-\frac{d}{dt}\,\frac{\partial L}{\partial \dot{\eta}_r}=-\omega_r^2\eta_r-\frac{d}{dt}\dot{\eta}_r=0
\end{equation}
or
\begin{equation}
\ddot{\eta}_r+\omega_r^2\eta_r=0\qquad{\rm for}\quad r=1\ldots N\,,
\end{equation}
which is a set of equations for $N$ decoupled harmonic oscillators with the
typical solutions
\begin{equation}\label{eq:normalcoordinatetimedep}
\eta_r(t)=\eta_r^+\,e^{i\omega_r t}+\eta_r^-\,e^{-i\omega_r t}\,.
\end{equation}
As before, the coefficients $\eta_r^+$ and $\eta_r^-$ need to be chosen to
meet initial conditions.

To summarize, the strategy of solving a problem of small oscillations around
the equilibrium of a coupled system of masses is as follows:
\begin{itemize}
\item Determine the mass matrix $\tensor{m}$ either directly if the
    coordinates are the Cartesian coordinates, or via
    (\ref{eq:coupledT}) and (\ref{eq:massmatrixelement}).
\item Find the elastic coupling matrix $\tensor{A}$ according to
  (\ref{eq:elasticitymatrix}).
\item Find the eigenfrequencies $\omega_r$ and a set of corresponding normalized
  eigenvectors $\vec{a}_r$ to the matrix $\tensor{m}^{-1}\cdot\tensor{A}$ for
  the normal modes of the system.
\item Find a combination of amplitudes $\alpha_r$ and phase shifts $\delta_r$
  for each eigenmode that satisfy the initial conditions via
  (\ref{eq:coupledsolution}) -- and you are done!
\end{itemize}
In practice, this strategy can be followed for relatively small systems with
not too many coordinates, because then the eigenvector search can be either
done manually, or very efficient numerical methods can be used. The difficulty
would be more in finding the elastic coupling matrix $\tensor{A}$ if
interaction between all masses take place. Examples for such problems are the
vibrations that can occur in molecules.

\subsection{Linear systems with next-neighbor couplings}
We now come to a generalization of the problem of coupled masses presented in
section \ref{ssec:twocoupledoscillators} from 2 to $N$ masses coupled to each
other. The coupling should be restricted to close neighbors, as it is typical
e.g.\ for lattice vibrations in a crystalline solid. However, the problem
is much more general, because in various areas of physics one is
restricted to such ``local'' interactions.

As the simplest example of such a problem, we consider again a linear chain
of masses that can move in one direction, coupled by springs:

\begin{figure*}[h]
\centerline{\includegraphics{manymasses.eps}}
\end{figure*}
The system looks similar for all masses, i.e.,  all masses are the same and
see a similar environment of neighbors and springs. The
equation of motion  is obtained in one of the usual ways, e.g.\ by considering
the total force $F_j$ acting on mass $j$:

\begin{figure*}[h]
\centerline{\includegraphics{couplingmodel.eps}}
\end{figure*}

\noindent Force $F_j$ is determined by position $u_j$ of mass $j$, and those of
its neighbors:
\begin{eqnarray}\label{eq:linearstringEOM}
F_j=m\ddot{u}_j&=&-k(u_j-u_{j-1})-k(u_j-u_{j+1})\nonumber\\
&=&k(u_{j-1}+u_{j+1}-2u_j)\,.
\end{eqnarray}
Index $j$ runs from 1 to $N$; we define $u_0\!:=\!0$ and
$u_{N+1}\!:=\!0$ to fulfill the boundary conditions at both ends of the chain
in expression (\ref{eq:linearstringEOM}).

A very similar problem appears when we consider a {\em transverse
  displacement} of masses from an equilibrium position of a coupled string
of masses. This approximates the situation of the string in a music
instrument. The string should have a static tension force $\tau$ in
equilibrium; the geometry of the problem is shown below:

\begin{figure*}[h]
\centerline{\includegraphics{stringdisplacement.eps}}
\end{figure*}

\noindent The transverse restoring force for mass $j$ is now given by
\begin{eqnarray}
F_j&=&-\tau\sin\phi-\tau\sin\phi'\nonumber\\
&\approx&-\tau(\tan\phi+\tan\phi')\nonumber\\
&=&-\tau\left(\frac{q_j-q_{j-1}}{d}+\frac{q_j-q_{j+1}}{d}\right)\nonumber\\
&=&\frac{\tau}{d}(q_{j-1}+q_{j+1}-2q_j)\,,
\end{eqnarray}
leading to the same equation of motion as (\ref{eq:linearstringEOM}) with a
different force coefficient:
\begin{equation}\label{eq:transversestringEOM}
m\ddot{q}_j=\frac{\tau}{d}(q_{j-1}+q_{j+1}-2q_j)\,.
\end{equation}
Using the usual harmonic ansatz $u_j=a_je^{i\omega t}$, we obtain the
algebraic equation (\ref{eq:eigenfrequencies}) for both the eigenfrequencies
and the eigenvectors; in components:
\begin{equation}\label{eq:lambdadef}
-a_{j-1}+\lambda a_j-a_{j+1}=0\quad{\rm for}\quad j=1\ldots N\,,\qquad{\rm with}\quad\lambda=(2-m\omega^2/k)\,,
\end{equation}
or in matrix form with $\tensor{D}=\tensor{A}-\omega^2\tensor{m}$:
\begin{equation}\label{eq:eigenmodeeq}
\tensor{D}\cdot\vec{a}=0\quad{\rm
  or}\quad
\left(\begin{array}{ccccc}
    \lambda&-1&0&0&\ldots\\
    -1&\lambda&-1&0&\ldots\\
    0&-1&\lambda&-1&\ldots\\
    &&\vdots&\ddots&
\end{array}\right)\cdot
\left(\begin{array}{c}a_1\\a_2\\\vdots\\a_N\end{array}\right)=0\,.
\end{equation}
To satisfy this matrix equation, the characteristic equation must hold:
\begin{equation}\label{eq:chareq2}
{\rm det}\,\tensor{D}=0\,,
\end{equation}
fixing the frequency $\omega$ of a solution.
We verify this for two special cases. For $N=1$, the
matrix $\tensor{D}$ reduces to the single value $\lambda$, and
(\ref{eq:chareq2}) simply becomes
\begin{equation}\lambda=0\,.\end{equation}
From this, with the definition of $\lambda$ in (\ref{eq:lambdadef}), we find
\begin{equation}\label{eq:freqsinglemass}
\omega=\sqrt{\frac{2k}{m}}\,.
\end{equation}
Compared to the result (\ref{eq:harmoscfreq}) for the harmonic oscillator, the
factor 2 here should not come as a surprise, because the we chose the boundary 
conditions such that the masses at the end are coupled to a wall with another
spring on each side:\vspace{2mm}

%\begin{figure*}[h!]
\centerline{\includegraphics{onemass.eps}}
%\end{figure*}

For $N=2$ we expect to reproduce the problem in section
\ref{ssec:twocoupledoscillators}. We have 
\begin{equation}
\tensor{D}=\left(\begin{array}{cc}\lambda&-1\\-1&\lambda\end{array}\right)\,,\qquad{\rm
  therefore}\quad{\rm det}\,\tensor{D}=\lambda^2-1=0\,,
\end{equation}
resulting in the solutions and corresponding eigenfrequencies seen in (\ref{eq:twofrequencies}):
\begin{equation}
\lambda_{1,2}=\pm1\,,\qquad{\rm and}\quad
\omega_{1,2}=\sqrt{\frac{2-\lambda_{1,2}}{m/k}}=\sqrt{\frac{2k\mp k}{m}} \,.
\end{equation}

One could come up with a recursive expression for the characteristic equation
for any $N$, but it is much more convenient to solve the problem differently.
\subsubsection{Solution with the wave approach}\label{ssec:waveapproach}
In the algebraic equation system (\ref{eq:lambdadef}), the structure for all
coefficients looks similar, with couplings only to next neighbors. We consider
the ansatz
\begin{equation}\label{eq:waveapproach}
a_j=ae^{i(j\gamma-\delta)}
\end{equation}
for the components of vector $\vec{a}$ in (\ref{eq:eigenmodeeq}), which is
eigenvector to $\tensor{m}^{-1}\cdot\tensor{A}$. The coefficients $a$ and
$\delta$ should be real values, describing a common amplitude and phase for all
components $a_j$, and $\gamma$ a phase shift between neighboring
components. The components $a_{j\pm1}$ for the next neighbors are  given by
\begin{equation}\label{eq:wavedef}
a_{j\pm1}=ae^{i([j\pm1]\gamma-\delta)}=e^{\pm i\gamma}a_j\,,
\end{equation}
which helps to simplify (\ref{eq:lambdadef}):
\begin{equation}
-a_{j-1}-a_{j+1}+\lambda a_j=0\quad\rightarrow\quad
(-e^{i\gamma}-e^{-i\gamma}+\lambda)a_j=0\quad\forall_{j=1\ldots N}\,.
\end{equation}
Since the amplitude $a_j$ is non-zero by construction (\ref{eq:waveapproach}),
the parenthesis has to vanish, which
leads to a condition between $\gamma$ and $\lambda$:
\begin{equation}
2\cos\gamma=\lambda=2-\frac{m}{k}\omega^2
\end{equation}
or
\begin{equation}\label{eq:phononeigenfreq}
\omega_r^2=\frac{k}{m}(2-2\cos\gamma_r)=\frac{2k}{m}(1-\cos\gamma_r)=\frac{4k}{m}\sin^2\frac{\gamma_r}{2}
\quad\rightarrow\quad
\omega_r=\sqrt{\frac{4k}{m}}\sin\frac{\gamma_r}{2}\,.
\end{equation}
There, the index $r$ indicates that we need to find $r$ eigenfrequencies, and
corresponding constants $\gamma_r$ and $\delta_r$.
We now need to fix the values of $\gamma_r$ and $\delta_r$ to meet the boundary
conditions. As we are looking for real-valued displacements, we could have
equally well chosen the real part of (\ref{eq:wavedef}), at the cost of more
complex expressions. We make this transition now:
\begin{equation}
a_{j,r}=a_re^{i(j\gamma_r-\delta_r)}\quad\rightarrow\quad
a_{j,r}=a_r\cos(j\gamma_r-\delta_r)\,.
\end{equation}
The boundary condition at the left end ($j=0$) requires that $a_{0,r}=0$ or
\begin{equation}
0=a_{0,r}=\cos(-\delta_r)\quad\Rightarrow\quad\delta_r=\frac{\pi}{2}\,,\qquad{\rm
  so}\quad a_{j,r}=a_r\sin(j\gamma_r)\,.
\end{equation}
The boundary condition on the right end of the chain requires
\begin{equation}
a_{N+1,r}=0\quad\Rightarrow \gamma_r(N+1)=s\pi\,,\quad s=1,2,3,\ldots
\end{equation}
Since we need $N$ different solutions for $\omega_r$, we simply choose $s=r$,
so
\begin{equation}
\gamma_r=\frac{r\,\pi}{N+1}\,,\quad r=1\ldots N\,.
\end{equation}
This leads to eigenvector components
\begin{equation}\label{eq:modestructure}
a_{j,r}=a_r\sin\left(j\frac{r\,\pi}{N+1}\right)
\end{equation}
and with (\ref{eq:phononeigenfreq}) to eigenfrequencies
\begin{equation}\label{eq:eigenfrequencies2}
\omega_r=\sqrt{\frac{4k}{m}}\sin\frac{r\,\pi}{2(N+1)}\,.
\end{equation}
As in (\ref{eq:normalcoord2}), we can express the individual deviations of the
masses via normal coordinates $\eta_r$ and the corresponding time dependence (\ref{eq:normalcoordinatetimedep}),
\begin{equation}
u_j(t)=\sum\limits_ra_{j,r}\Big[\eta_r^+e^{i\omega_rt}\,+\,\eta_r^-e^{-i\omega_rt}\Big]\,,
\end{equation}
with adequate choices for the $\eta_r^\pm$ to meet initial conditions if
required.
\subsubsection{Resulting mode structure}
The mode structure contained in expression (\ref{eq:modestructure}) resembles
that of  standing waves with different wavelengths; we visualize the different
eigenvector components $a_{j,r}$ for the example with $N=3$ in the graph
below. The eigenmodes for $r=1\ldots3$ follow the pattern of {\em standing
  waves}, sampled at discrete points. The standing wave meets the boundary
conditions at the auxiliary positions $j=0$ and $j=N+1=4$.

\begin{figure*}[h]
\centerline{\includegraphics{modestructure.eps}}
\end{figure*}
Modes for $r=4$ and $r=8$ do not lead to a meaningful
oscillations, as all eigenvector components vanish. The modes corresponding to
$r=5,6,7$ reproduce the modes structure corresponding to $r=3,2,1$, with an
additional minus sign for all components. Hence, they do not constitute any
new mode, and the description of modes with $r=1,2,3$ is indeed complete.

\subsubsection{Dispersion relation}
Since the amplitude distribution $a_{j,r}$ for the different modes $r$ follow
a standing wave pattern, we can use the wavelength of this wave to
characterize the mode. For this, we assume that all masses spaced by the same
distance $d$ from their next neighbors. Then, the total length of
the chain from endpoint to endpoint is $L=(N+1)d$. The wavelength of the
fundamental mode ($r=1$) is then $\Lambda_1=2L$, and from
(\ref{eq:modestructure}) the  wavelength of a higher order mode
can be written as
\begin{equation}
\Lambda_r=\frac{2L}{r}\,.
\end{equation}
One then can identify mass $j$ also by its distance
$x_j=jd$ from the left boundary corresponding to $j=0$, and express the
displacement $u_j$ in that mode by
\begin{eqnarray}\label{eq:spatialmode}
a_{j,r}&=&a_r\sin\left(j\frac{r\pi}{N+1}\right)
=a_r\sin\left(\frac{(dj)r\pi}{L}\right)\nonumber\\
&=&a_r\sin\left(x_j\frac{2L}{\Lambda_r}\frac{\pi}{L}\right)
= a_r\sin\left(x_j\frac{2\pi}{\Lambda_r}\right)=a_r\sin\left(x_j\,q_r\right)\,.
\end{eqnarray} 
Here, $q_r$ is a {\em wave number} and indicates a {\em spatial frequency}:
\begin{equation}
q_r=\frac{2\pi}{\Lambda_r}=\frac{2\pi r}{2L}=\frac{r\pi}{(N+1)d}\,.
\end{equation}
Its maximal value is
\begin{equation}
q_{r,{\rm
    max}}=q_N=\frac{\pi}{d}\frac{N}{N+1}\,,
\end{equation}
which converges to $q_{\rm max}=\pi/d$ for large $N$. This is the typical
situation e.g. for the motion of the atoms in a crystal, with lattice spacing
$d$. There, the wave number $q$ is a much better indicator for
the mode as the integer mode index $r$ because of the large number of atoms involved. The upper limit $q_{\rm max}$
corresponds to a minimal wavelength of $\Lambda_{\rm min}=2\pi/q_{\rm
  max}=2d$; a wavelength smaller than twice the lattice
spacing $d$ is not meaningful for a description of modes.

In the same way, the dependency of the eigenfrequencies $\omega_r$ in
(\ref{eq:eigenfrequencies2}) from mode index $r$ can be described by the wave
number $q$: 
\begin{eqnarray}\label{eq:dispersionrelation}
\omega(q)&=&\sqrt{\frac{4k}{m}}\sin\frac{r\pi}{2(N+1)}\nonumber\\
&=&\omega_{\rm
  max}\sin\frac{qd}{2}\,,\qquad{\rm with}\quad \omega_{\rm max}=\sqrt{\frac{4k}{m}}\,.
\end{eqnarray}
A function that relates the oscillation frequency of a mode with its
wavenumber is called {\em dispersion relation} for a coupled system. The
figure below shows the dispersion relation for a coupled linear chain of
masses with $N=10$:

\begin{figure*}[h]
\centerline{\includegraphics{dispersionrelation.eps}}
\end{figure*}
The maximal frequency $\omega_{\rm max}$ corresponds to an oscillation mode
where adjacent lattice sites oscillate in opposite direction; thus, the spring
constant, compared to the case of a single mass subject to two springs
attached to a fixed wall, effectively doubles compared to the result
(\ref{eq:freqsinglemass}).

The theory of coupled spring/mass systems is important in solid state
physics; the problem above is only the simplest example of a lattice vibration
in crystalline solids. It has to be extended to three dimensions, and typically,
more complex crystal structures with more than one atom in a crystal unit cell
need to be considered. But the basic treatment is the same as outlined in this
section: oscillation modes are indexed by a wave number (or wave vector, in a
three-dimensional case), and other parameters that indicate transverse or
longitudinal displacements. As the frequencies of oscillation can be quite
high in a solid, such oscillations need to be treated quantum mechanically,
giving rise to ``phonons'' as quasi-particles corresponding to a particular
mode index $q$. However, the whole dispersion relation of lattice vibrations
is a purely classical mechanics problem.

\subsection{Transition to a continuum}
In the last section, we already replaced the particle index $j$ by
a position index $x_j$, and the discrete eigenmode index $r$ by a wave
number $q$. By increasing the number $N$ of masses, while shrinking their
distance $d$, but keeping the overall physical properties of the string,
like its length $L$, total mass $M$ and elastic properties fixed,
we will arrive at a continuous distribution of masses and springs. This is a
suitable description of solids, because it is often neither possible nor
interesting to keep track of the position of all its constituting masses.

We first consider the basic variables
of the problem. In the discrete case, these are the displacements
$u_j$, which can be expressed by time-dependent normal coordinates
$\eta_r$ via (\ref{eq:normalcoord2}), (\ref{eq:spatialmode}):
\begin{equation}\label{eq:continuumsolution}
u_j(t)=\sum\limits_r\eta_r(t)a_{j,r}=\sum\limits_r\eta_r(t)\sin(x_jq_r)\,.
\end{equation}
The $\eta_r$ exhibit the simple harmonic
oscillator dynamics (\ref{eq:normalcoordinatetimedep}) at frequency
$\omega_r$. Expression (\ref{eq:continuumsolution}) has already a form to make
the transition 
\begin{equation}
u_j(t)\rightarrow u(x,t)\,,
\end{equation}
to a function with a continuous position variable $x$. The time
dependency of $u$ for a given mode $q$ still is a harmonic
oscillation, but the $\omega(q)$ changes for
$d\rightarrow0$. For small $d$, we approximate the sinus in dispersion
relation (\ref{eq:dispersionrelation}) for small angles:
\begin{eqnarray}\label{eq:continuousdispersion}
\omega(q)&=&\sqrt{\frac{4k}{m}}\sin\frac{qd}{2}\nonumber\\
&\approx&\sqrt{\frac{4k}{m}}\,\frac{qd}{2}
\,=\,q\,\sqrt{\frac{d}{m}}\,\sqrt{kd}\,.
\end{eqnarray}
The result is written
such that the first root is a ratio of distance over mass. If the global
properties of the problems stay fixed, this ratio is the inverse of a linear
mass density $\rho$. The second root contains a product of a spring constant
$k$ and distance $d$, which also remains constant for the limit
$d\rightarrow0$:  if a spring made out of a homogenous elastic 
material is cut in two pieces, an application of the same force to the short
piece will lead to {\em half} the compression or
extension of the original spring --- to compress it to the same length,
one would need to apply twice the force, i.e., the product 
of length $d$  and spring constant $k$  will not change. This product
\begin{equation}
K:=kd
\end{equation}
describes the stiffness of the material, and can be calculated from the elastic
modulus or \href{https://en.wikipedia.org/wiki/Young's_modulus}{Young's
  modulus}\footnote{Named after
  \href{https://en.wikipedia.org/wiki/Thomas_Young_(scientist)}{Thomas Young},
  1773-1828, but discovered around 1727 by
  \href{https://en.wikipedia.org/wiki/Leonhard_Euler}{L. Euler}.} $E$ of a
material via $K=EA_0$ for a given cross section $A_0$. Dispersion relation
(\ref{eq:continuousdispersion}) then becomes 
\begin{equation}
\omega(q)=q\,\sqrt{\frac{K}{\rho}}\,,
\end{equation}
which is linear in $q$ and has no upper limit for $q$ or $\omega$.

Similarly, a continuum relation for the transverse displacement for a string
under a tensional force $\tau$ can be calculated: the equations of motion
(\ref{eq:linearstringEOM}) for longitudinal, and
(\ref{eq:transversestringEOM}) for a transverse displacement are essentially
the same. All results for the longitudinal case can be transferred to the
transverse case by replacing $k$ with $\tau/d$, changing the
continuous dispersion relation (\ref{eq:continuousdispersion}) to
\begin{equation}
\omega(q)=q\,\sqrt{\frac{d}{m}}\,\sqrt{\frac{\tau}{d}\,d}=q\,\sqrt{\frac{\tau}{\rho}}\,.
\end{equation}

\subsubsection{Coupled equations of motion in a continuum}
Instead of transferring the results from a discrete problem to a continuum,
one can also move from a discrete set of coupled equations of motion like
(\ref{eq:linearstringEOM}) or (\ref{eq:transversestringEOM}) to a continuum
description, and solve the problem directly. For this, we rewrite
(\ref{eq:linearstringEOM}) to identify the continuum properties:
\begin{equation}\label{eq:waveequation0}
m\ddot{u}_j=k(u_{j+1}+u_{j-1}-2u_j)\quad\rightarrow\quad
\frac{m}{d}\ddot{u}_j=(kd)\left(\frac{u_{j+1}+u_{j-1}-2u_j}{d^2}\right)
\end{equation}
The last parenthesis approximates the second derivative of $u$ with respect
to the position $x_j$. For $d\rightarrow0$, 
\begin{equation}
\frac{m}{d}\rightarrow\rho\,,\quad k\,d\rightarrow K\,,\quad 
\frac{u_{j+1}+u_{j-1}-2u_j}{d^2}\rightarrow\frac{\partial^2 u(x)}{\partial x^2}\,,
\end{equation}
so that the equation of motion (\ref{eq:waveequation0}) becomes
\begin{equation}\label{eq:waveequation}
\rho\frac{\partial^2 u}{\partial t^2}=K\frac{\partial^2 u(x)}{\partial
  x^2}\qquad{\rm or}\qquad
\frac{\partial^2 u(x,t)}{\partial t^2}-\frac{K}{\rho}\frac{\partial^2
  u(x,t)}{\partial  x^2}=0\,.
\end{equation}
This is a {\em partial differential equation} that mixes differentiations
with respect to the different continuous variables $x$ and $t$ of the function
$u(x,t)$. The specific form of (\ref{eq:waveequation}) is called \href{https://en.wikipedia.org/wiki/Wave_equation}{\em
  wave equation}, and has a particular set of solutions.

\subsubsection{Solving the wave equation by separation of variables}
One way of solving the wave equation is to assume a harmonic time dependency
that is multiplied with some space-dependent part:
\begin{equation}
u(x,t)=v(x)e^{i\omega t}\,.
\end{equation}
Inserting this into the wave equation leads to
\begin{equation}
\left[-\omega^2 v(x)-\frac{K}{\rho}\frac{\partial^2
  v(x)}{\partial  x^2}\right]e^{i\omega t}=0\,,
\end{equation}
where one can divide by the time-dependent oscillation, and end up with
\begin{equation}\label{eq:Helmholtz}
\frac{\partial^2  v(x)}{\partial  x^2}+q^2 v(x)=0\,,\quad{\rm with}\quad
q^2=\frac{\rho \omega^2}{K}\,.
\end{equation}
This is an ordinary differential equation (or a simpler partial differential
equation if more than one space dimension is involved), and is referred to as
the \href{https://en.wikipedia.org/wiki/Helmholtz_equation}{\em Helmholtz
  equation}\footnote{after
  \href{https://en.wikipedia.org/wiki/Hermann_von_Helmholtz}{Hermann von
    Helmholtz}, 1821-1894}. For one dimension $x$, it has the same structure
as the equation of motion of a harmonic oscillator, which is consistent with
finding sinusoidal solutions for the spatial structure of eigenmodes in
(\ref{eq:continuumsolution}). The $q$ in expression (\ref{eq:Helmholtz}) is
exactly the wave number we used before, and we can directly extract the
dispersion relation.

The observation of continuum solutions can actually help to motivate solutions
for discrete variable cases; in section \ref{ssec:waveapproach}, the wave
approach was presented without a reasoning, but the solution of a continuum
problem really motivated the choice in (\ref{eq:waveapproach}).

\subsubsection{Propagating solutions to the wave equation}
Apart from solutions (\ref{eq:continuumsolution}) we derived from a finite
number of coupled masses, or could have gotten out of the Helmholtz equation
(\ref{eq:Helmholtz}), the wave equation (\ref{eq:waveequation}) has
 other interesting solutions. We try the the  ansatz
\begin{equation}
u(x,t)=w(x\pm vt)\,,
\end{equation}
i.e., we replace the function of two variable by a function $w$ of a single
variable that is a combination of $x$ and $t$. To check if this function can
solve the wave equation, we need derivatives of $u$ with respect to $t$ and
$x$: 
\begin{eqnarray}
\frac{\partial^2 u}{\partial x^2}&=&\frac{\partial}{\partial x}\big[w'(x\pm
vt)\big]\,=\,w''(x\pm vt)\,,\nonumber\\
\frac{\partial^2 u}{\partial t^2}&=&\frac{\partial}{\partial t}\big[\pm v\, w'(x\pm vt)\big]=v^2w''(x\pm vt)\,,
\end{eqnarray}
where $w''$ indicates the second derivative of function $w$ with respect to
its parameter. Inserting those derivatives into the wave equation leads to
\begin{equation}
v^2w''(x\pm vt)-\frac{K}{\rho}w''(x\pm vt) =
\left(v^2-\frac{K}{\rho}\right)w''(x\pm vt)=0
\end{equation}
For $v^2=K/\rho$, this equation can be fulfilled for {\em any} function $w$,
as long as it is differentiable. This is a remarkable result: Any initial
distribution $u(x,t=0)=w(x)$ is supported. Depending on the initial
conditions, this 
distribution propagates either in positive or negative direction (or a
combination of both) with a velocity 
\begin{equation} 
v=\sqrt{\frac{K}{\rho}}\,,\qquad{\rm or}\quad v=\sqrt{\frac{\tau}{\rho}}
\quad{\rm for\,\,transverse\,\,displacements\,.}
\end{equation}
This velocity is therefore the {\em speed of sound} in a solid, as sound is
the phenomenon of local displacements that propagate through material via
elastic coupling.

\subsubsection{Derivation of the wave equation via the Hamilton principle}
While the transition from a discrete mass problem to a continuum in the last
section leads to the correct wave equation of motion in an elastic
one-dimensional continuum, the equation can also be derived via the
Euler-Lagrange mechanism. The details are beyond the scope of this course, so
we only cover this approach very briefly.

The basis for the Euler-Lagrange formalism is the knowledge of the total
kinetic and total potential energy. We can make the transition from the total
kinetic energy of the discrete chain to a continuum in a similar way as in the
previous chapter:
\begin{equation}
T=\frac{1}{2}\sum\limits_jm\dot{u}_j^2=\frac{1}{2}\,\frac{m}{d}\sum\limits_j\dot{u}_j^2d\,.
\end{equation}
The ratio $m/d$ is again the linear mass density $\rho$. The sum of over all
velocities with the distance $d$ can be replaced by an integral over the length
of the string, leading for $d\rightarrow0$ to the asymptotic expression for
the total kinetic energy
\begin{equation}
T=\frac{1}{2}\rho\int\limits_0^L\left(\frac{\partial u(x,t)}{\partial t}\right)^2\,dx\,.
\end{equation}
For the potential energy, we do a similar transition:
\begin{equation}
U=\frac{1}{2}k\sum\limits_j(u_{j+1}-u_{j})^2=\frac{1}{2}\,(kd)\sum\limits_j\left(\frac{u_{j+1}-u_j}{d^2}\right)^2\,d\,.
\end{equation}
The fraction in the sum (with the distance $d$) will become the partial derivative of the displacement
with respect to $x$ for $d\rightarrow0$, and the sum will go over into an
integral:
\begin{equation}
U=\frac{1}{2}K\int\limits_0^L\left(\frac{\partial u(x,t)}{\partial x}\right)^2\,dx\,.
\end{equation}
Together with the kinetic energy $T$, this leads to a Lagrange function
\begin{equation}\label{eq:LagrangeCont}
L=T-U=\int\limits_0^L\left[\frac{\rho}{2}\left(\frac{\partial u}{\partial
      t}\right)^2-\frac{K}{2}\left(\frac{\partial u}{\partial
      x}\right)^2\right]\,dx=\int\limits_0^L{\cal L}\left(u,\frac{\partial
    u}{\partial t},\frac{\partial u}{\partial x}\right)\,dx
\end{equation}
for the continuum, with a {\em Lagrange density} $\cal{L}$ that depends on the
function $u(x,t)$, and its first partial derivatives with respect to $x$ and
$t$.

Without further proof, the equivalent to the Euler-Lagrange equation
(\ref{eq:eulerlagrange}) for problems with two continuous parameters $x$ and $t$
is
\begin{equation}\label{eq:ELagContinuous}
\frac{\partial \cal{L}}{\partial u}\,-\,\frac{\partial}{\partial t}\left[
\frac{\partial \cal{L}}{\partial \left(\frac{\partial u}{\partial t}\right)}
\right]\,-\,\frac{\partial}{\partial x}\left[
\frac{\partial \cal{L}}{\partial \left(\frac{\partial u}{\partial x}\right)}\right]=0\,.
\end{equation}
The Lagrange density for the elastically coupled mass density in
(\ref{eq:LagrangeCont}) 
\begin{equation}
{\cal L}\left(u,\frac{\partial u}{\partial t},\frac{\partial u}{\partial
    x}\right) =
\frac{\rho}{2}\left(\frac{\partial u}{\partial
      t}\right)^2-\frac{K}{2}\left(\frac{\partial u}{\partial
      x}\right)^2\,,
\end{equation}
has partial derivatives
\begin{equation}
\frac{\partial \cal{L}}{\partial \left(\frac{\partial u}{\partial t}\right)}
=\rho\,\frac{\partial u}{\partial t}\,,\qquad{\rm and}\quad
\frac{\partial \cal{L}}{\partial \left(\frac{\partial u}{\partial x}\right)}
=-K\,\frac{\partial u}{\partial x}\,.
\end{equation}
Inserting those into the Euler-Lagrange equation (\ref{eq:ELagContinuous}),
and observing that $\cal{L}$ does not explicitly depend on $u$ (so $\partial
{\cal{L}}/\partial u=0$) leads to
\begin{equation}
-\frac{\partial}{\partial t}\left[\rho\,\frac{\partial u}{\partial
    t}\right]\,-\,
\frac{\partial}{\partial x}\left[-K\,\frac{\partial u}{\partial x}\right]=
-\rho\,\frac{\partial^2 u}{\partial t^2}+K\,\frac{\partial^2 u}{\partial x^2}=0\,,
\end{equation}
which reproduces the wave equation (\ref{eq:waveequation}). 

It should be stated that mechanical problems are usually never treated with
this mechanism, since the formalism is way too complicated, and the respective
partial differential equations for the displacement field $u(x,t)$ can be
obtained in a much simpler way. The method outlined above, 
however, is used in high energy physics, and when dealing with interactions
that do not easily lead to field equations otherwise.

\section{Non-inertial reference frames}
So far, we have described mechanical problems only in inertial reference
frames where Newton's first law holds, stating 
that a body not subject to a force remains in a state of uniform motion.
However, there are important cases where inertial reference frames are
inadequate or complicated. Notably, specifying the motion of bodies on the
surface of the Earth is such an example. Because of the rotation of the
Earth, Newton's laws (or equivalently, the equations of motion derived via the
Euler-Lagrange formalism) do only hold approximately. In this section, we
discuss a suitably modified version of Newton's second law in these systems.

\subsection{Coordinate transformation}
Consider two reference frames $F$ and
$F'$ that are moving with respect to each other, and may also change their
respective orientation over time:

\begin{figure*}[h]
\centerline{\includegraphics{twoframes0.eps}}
\end{figure*}

The transformation at any instant can be separated in a translation between the
two origins $O$ and $O'$, and a single rotation that adjusts the orientation
of the coordinate systems with respect to each other (see section
\ref{ssec:coordinatetransformations}). If $\vec{x}$ is a vector described
by coordinates $x_i$ in $F$ , and by coordinates $x'_i$ in
$F'$, i.e.\
\begin{equation}
\vec{x}=\sum\limits_{i=1}^3x_i\vec{e}_i\,=\,\sum\limits_{i=1}^3x'_i\vec{e'}_i\,,
\end{equation}
then the two coordinate representations are connected via a rotation matrix
$\tensor{R}$
\begin{equation}\label{eq:rotatevector1}
\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)=
\tensor{R}\cdot\left(\begin{array}{c}x'_1\\x'_2\\x'_3\end{array}\right)\,.
\end{equation}
As seen in section \ref{ssec:coordinatetransformations}, the rotation around
the $x_3$ axis is e.g.\ represented by
\begin{equation}
\tensor{R}_3(\phi)=\left(\begin{array}{ccc}
    \cos\phi&-\sin\phi&0\\\sin\phi&\cos\phi&0\\0&0&1
\end{array}\right)\,.
\end{equation}
We then made the transition to infinitesimal rotations, e.g. for the $x_3$ axis
\begin{equation}
\tensor{R}_3(\phi)\rightarrow\tensor{R}_3(d\phi)=\tensor{1}+\tensor{\epsilon}_3(d\phi)=\tensor{1}+
\left(\begin{array}{ccc}
    0&-d\phi&0\\d\phi&0&0\\0&0&0 \end{array}\right)\,, 
\end{equation}
with the unity matrix $\tensor{1}$. As shown in section
\ref{ssec:smallrotations}, infinitesimal rotations around different axes can
simply be added up, leading to the general infinitesimal correction matrix
\begin{equation}
\tensor{\epsilon}(d\phi_1, d\phi_2,d\phi_3)=
\left(\begin{array}{ccc}
0&-d\phi_3&d\phi_2\\ d\phi_3&0&-d\phi_1\\-d\phi_2&d\phi_1&0
\end{array}\right)\,,
\end{equation}
where the $d\phi_i$ describe infinitesimal rotations around coordinate
axes $i$. If all infinitesimal rotations $d\phi_i$
are combined into a vector $d\vec{\phi}=(d\phi_1,d\phi_2,d\phi_3)$, the action of
correction matrix $\tensor{\epsilon}$ can be expressed as a vector product:
\begin{equation}
\tensor{\epsilon}\cdot\vec{x}=d\vec{\phi}\times\vec{x}
\end{equation}
Therefore, we can rewrite the right side of (\ref{eq:rotatevector1}) as
\begin{equation}\label{eq:differentialrot2}
\tensor{R}\cdot\vec{x}=(\tensor{1}+\tensor{\epsilon})\cdot\vec{x}
=\vec{x}+\underbrace{d\vec{\phi}\times\vec{x}}_{=:d\vec{x}}
\end{equation}
This means that for an infinitesimal rotation $d\vec{\phi}$, the coordinate
vector needs to be corrected by the infinitesimal amount $d\vec{x}$ to make
the transition between two coordinate systems.

\subsubsection{Transformation of time derivatives of vectors}
Dividing the differential $d\vec{x}$  in (\ref{eq:differentialrot2}) by
an infinitesimal time $dt$ leads to an expression for the temporal derivative
of vector $\vec{x}$: 
\begin{equation}\label{eq:timederivativebasevec}
\frac{d\vec{x}}{dt}=\frac{d\vec{\phi}}{dt}\times\vec{x}=\vec{\omega}\times\vec{x}\,,
\end{equation}
where $\vec{\omega}$ is the instantaneous angular velocity. This expression
reproduces the relation (\ref{eq:rotationspeed}) between the instant velocity
$\vec{v}=d\vec{r}/dt$ and the position $\vec{r}$ introduced earlier, but holds
for {\em any} vector $\vec{x}$, and allows evaluating the temporal
derivative of a vector due to a rotation of the reference frame.

%\begin{figure*}[h]
%\centerline{\includegraphics{twoframes_e.eps}}
%\end{figure*}


We now consider the temporal derivative of a vector $\vec{x}$ expressed as a
linear combination of base vectors $\vec{e'}_i$ of $F'$,
\begin{equation}\label{eq:aprimedef}
\vec{x}(t)=\sum\limits_ix'_i(t)\,\vec{e'}_i(t)\,.
\end{equation}
Both the coefficients $x'_i$ and the basis vectors $\vec{e'}_i$ are
time dependent, so
\begin{equation}
\left.\frac{d\vec{x}}{dt}\right|_F
=\frac{d}{dt}\sum\limits_ix'_i\,\vec{e'}_i
=\sum\limits_i\left(\dot{x}'_i\,\vec{e'}_i\,+\,x'_i\,\vec{\dot{e}'}_i\right)
=:\left.\frac{d\vec{x}}{dt}\right|_{F'}+\sum\limits_ix'_i\,\vec{\dot{e}'}_i
\end{equation}
The index $F$ indicates that the derivative is supposed to be taken in frame
$F$, capturing both the time derivative of the coordinates $x'_i$ and the base
vectors $\vec{e'}_i$. The index $F'$ in the temporal derivative of $\vec{x}$
indicates that this is the derivative in system $F'$: only the temporal
derivatives of the coefficients $x'_i$ are taken, multiplied with the base
vectors $\vec{e'}_i$ that do not change in time in reference system $F'$. The
second term contains temporal derivatives of $\vec{e'}_i$, as seen from
reference frame $F$. For this, we can use
(\ref{eq:timederivativebasevec}):
\begin{equation}
\vec{\dot{e}'}_i=\frac{d\vec{e'}_i}{dt}=\vec{\omega}\times\vec{e'}_i\,,
\end{equation}
With this, we can continue with the time derivative of $\vec{x}$:
\begin{eqnarray}\label{eq:timederivativerot}
\left.\frac{d\vec{x}}{dt}\right|_F&=&\left.\frac{d\vec{x}}{dt}\right|_{F'}
\,+\,\sum\limits_ix'_i\,(\vec{\omega}\times\vec{e'}_i)\nonumber\\
&=&\left.\frac{d\vec{x}}{dt}\right|_{F'}\,+\,
\vec{\omega}\times\sum\limits_ix'_i\vec{e'}_i\,=\,
\left.\frac{d\vec{x}}{dt}\right|_{F'}
\,+\,\vec{\omega}\times\vec{x}\,.
\end{eqnarray}
The temporal derivative of a vector $\vec{x}$ as seen from reference frame $F$
is therefore given by the temporal derivative of the vector as seen in
reference frame $F'$, 
\begin{equation}
\left.\frac{d\vec{x}}{dt}\right|_{F'}=\sum\limits_i\dot{x}'_i\,\vec{e'}_i\,,
\end{equation}
corrected by a term that takes into account the time
evolution of the base vectors $\vec{e'}_i$ of frame $F'$ as seen from frame $F$.

\subsubsection{Transformation of velocities and accelerations}
We now consider the transformation of the characteristic vectors of a moving
point $P$ between two reference frames $F$ and $F'$. The origin $O'$ of $F'$
should be displaced by a vector $\vec{R}$ from the origin of $F$:

\begin{figure*}[h]
\centerline{\includegraphics{twoframes2_e.eps}}
\end{figure*}

\noindent The position vector $\vec{r'}$ of point $P$  with respect to origin
$O'$ can be expressed by vectors in system $F$:
\begin{equation}
\vec{r}=\vec{R}+\vec{r'}\,.
\end{equation}
The displacement vector $\vec{R}$ between the two reference frames may also
change over time. Therefore, the velocity of $P$ as seen from frame $F$ is
given by
\begin{eqnarray}\label{eq:velocitytransformation}
\frac{d\vec{r}}{dt}&=&\frac{d\vec{R}}{dt}\,+\,\left.\frac{d\vec{r'}}{dt}\right|_F\nonumber\\
&=&\frac{d\vec{R}}{dt}\,+\,\left.\frac{d\vec{r'}}{dt}\right|_{F'}\,+\,
\vec{\omega}\times\vec{r'}\,,
\end{eqnarray}
where the indices $F$ and $F'$ indicate the system with which respect a
temporal derivative is taken. The above expression can be rewritten as
\begin{equation}
\vec{v}=\vec{V}+\vec{v'}_{\!F'}+\vec{\omega}\times\vec{r'}\,,
\end{equation}
where $\vec{V}=\dot{\vec{R}}$ is the velocity of origin $O'$ in $F$, the
velocity 
$\vec{v'}_{\!F'}$ of $P$ is the one seen in system $F'$ with respect to the
origin $O'$, angular velocity $\vec{\omega}$ captures the instantaneous change
of orientation of the two coordinate systems.

To obtain the acceleration of point $P$, as seen from reference system $F$, we
take another temporal derivative of 
(\ref{eq:velocitytransformation}):
\begin{eqnarray}\label{eq:accelerationtransformation}
\vec{a}&=&\frac{d\vec{V}}{dt}\,+\,
\left.\frac{d\vec{v'}_{\!F'}}{dt}\right|_F\,+\left[\,\dot{\vec{\omega}}\times\vec{r'}\,+\,
\vec{\omega}\times\left.\frac{d\vec{r'}}{dt}\right|_F\right]\nonumber\\
&=&\ddot{\vec{R}}\,+\,
\left(\left.\frac{d\vec{v'}_{\!F'}}{dt}\right|_{F'}+\vec{\omega}\times\vec{v'}_{\!F'}\right)\,+\,
\Bigg[\dot{\vec{\omega}}\times\vec{r'}\,+\,
\vec{\omega}\times\Bigg(\underbrace{\left.\frac{d\vec{r'}}{dt}\right|_{F'}}_{=\vec{v'}_{\!F'}}\,+\,\vec{\omega}\times\vec{r'}\Bigg)\Bigg]\nonumber\\
&=&\ddot{\vec{R}}\,+\,\vec{a'}_{\!F'}\,+\,
2\vec{\omega}\times\vec{v'}_{\!F'}+\dot{\vec{\omega}}\times\vec{r'}\,+\,
\vec{\omega}\times(\vec{\omega}\times\vec{r'})
\end{eqnarray}
In the first line, only the time dependency of $\vec{\omega}$ and $\vec{r'}$
was considered. From the first to the second line, the temporal derivatives of
$\vec{v'}$ and $\vec{r'}$ were corrected for the rotation of the two reference
frames with respect to each other. The newly introduced term $\vec{a'}_{\!F'}$
is the acceleration of $P$, as seen in reference frame $F'$:
\begin{equation}
\vec{a'}_{\!F'}=\sum\limits_i\ddot{r}'_i\vec{e'}_i\,.
\end{equation}

\subsection{Dynamics in non-inertial reference frames}
So far, we only considered the kinematics of transformations between
reference frames $F$ and $F'$. We are not yet able to come up with a
description of the dynamics of a system, as it was defined by Newton's second
law.

To accomplish this, we now postulate that $F$ is an inertial reference frame
for which Newton's laws apply, while $F'$ can be a non-inertial reference frame.
The dynamics of a mass point $m$ can then be described in the inertial
reference frame $F$ via Newton's second law (\ref{eq:ForceDefinition}):
\begin{equation}
\vec{F}=m\,\vec{a}\,,
\end{equation}
where the force $\vec{F}$ is a sum of all interactions with the environment or other
masses, like gravitation, Coulomb interaction etc. Since there is also a
meaningful acceleration vector $\vec{a'}_{\!F'}$ in the non-inertial reference
frame $F'$, we {\em define} an {\em effective force} $\vec{F}_{\rm eff}$ that
allows writing down the equivalent to Newton's second law in the
non-inertial reference frame:
\begin{equation}
\vec{F}_{\rm eff}:=m\,\vec{a'}_{\!F'}\,.
\end{equation}
With the expression (\ref{eq:accelerationtransformation}) for transforming the
accelerations between $F$ and $F'$, we find
\begin{equation}\label{eq:inertialforces}
\vec{F}_{\rm eff}=\vec{F}-m\ddot{\vec{R}}-m\dot{\vec{\omega}}\times\vec{r'}
-2m\vec{\omega}\times\vec{v'}_{\!F'}-m\vec{\omega}\times(\vec{\omega}\times\vec{r'})
\end{equation}
This means that in the non-inertial reference frame, the effective force
$\vec{F}_{\rm eff}$ contains not only the ``true'' forces generated by
interaction of masses, charges etc also seen in the inertial reference frame
$F$, but also a number of so-called {\em inertial forces} that are a
consequence of $F'$ not being an inertial reference frame. 

The first term results from an acceleration of the reference frame with
respect to the inertial frame; this is the apparent force one observes in an
accelerating elevator, or in an accelerating/decelerating vehicle. The second
term is due to an angular acceleration, and has perhaps not an obvious
presence in everyday life.

The third term is an apparent force that is proportional to the velocity
$\vec{v'}_{\!F'}$  in the non-inertial reference frame $F'$, and referred to
as {\em Coriolis force}.

The last term in (\ref{eq:inertialforces}) is the {\em centrifugal force} that
a body in reference frame $F'$ feels that is proportional to the square of the
angular velocity.

\subsubsection{Centrifugal force}
We first look at the geometry of the centrifugal force term in (\ref{eq:inertialforces}),
\begin{equation}
\vec{F}_{\rm centrif.}=-m\vec{\omega}\times(\vec{\omega}\times\vec{r'})\,.
\end{equation}

\noindent \raisebox{-1.6cm}{\includegraphics{centrifugal.eps}}\hfill
\begin{minipage}{8cm}
The orientation of the centrifugal force can be seen in the figure: For an
origin $O'$ of $F'$ on the rotation axis, the vector
$\vec{\omega}\times\vec{r'}$ is the velocity vector tangential
to the trajectory of the point $P$. The resulting second vector product with
the angular velocity $\vec{\omega}$ makes the centrifugal force pointing
radially away from the rotation axis. 
\end{minipage}\vspace{1mm}

\noindent The modulus of the centrifugal force is given by
\begin{equation}
\left|\vec{F}_{\rm centrif.}\right|=m\,\omega^2\left|\vec{r'}\right|\sin\alpha=m\,\omega^2R\,,
\end{equation}
where $R$ is the shortest distance of $P$ from the rotation axis, replicating
the well-known expression for the centrifugal force.

The centrifugal force is e.g. responsible for the deviation of the shape of
the Earth from an ideal sphere; on the equator, the distance from the center
of the earth is about 21.4km larger than on the poles due to the daily
rotation. Furthermore, the
direction of the local acceleration e.g. felt by a mass on a string does not
point directly towards the center-of-mass of the earth, but must be corrected
to take care of the centrifugal term.


\subsubsection{Coriolis force}
 The Coriolis force term\footnote{described by
\href{https://en.wikipedia.org/wiki/Gaspard-Gustave_de_Coriolis}{Gaspard-Gustave
 de Coriolis}, 1792-1843; published in J. de l'Ecole royale polytechnique {\bf
 15}, 144--154 (1835)} in (\ref{eq:inertialforces})
\begin{equation}\label{eq:Coriolisforce}
\vec{F}_{\rm Coriolis}=-2m\vec{\omega}\times\vec{v'}_{\!F'}
\end{equation}
is only present if a body moves with respect to the moving reference system,
i.e., $\vec{v'}_{\!F'}\neq0$.
The geometry of the Coriolis force can be easily seen for a mass point that
moves with a velocity $\vec{v'}$ on a rotating platform:

\begin{figure*}[h]
\centerline{\includegraphics{rotatingplatform.eps}}
\end{figure*}
In the left figure, the mass point is moving radially towards the center; in a
coordinate system attached to the rotating platform, the velocity $\vec{v'}$
is parallel to the $x'_2$ direction. The resulting Coriolis force according to
(\ref{eq:Coriolisforce}) points in the $x'_1$ direction. This can be
understood in terms of an inertial effect: when moving radially in, the mass
point has an angular momentum that would be too large for a new (static)
position at a smaller radial distance. The mass point tries to retain its
momentum, which appears as an accelerating force in the tangential
direction in the rotating system. In the right figure, the velocity $\vec{v'}$
is along the $x'_1$ direction, resulting in a Coriolis force pointing radially
away from the rotation axis in the rotating frame. This can be interpreted as
an additional centrifugal term, because with its additional tangential
velocity, it appears to have a larger angular velocity than the rotating
reference frame $F'$.

The Coriolis force has consequences on moving bodies on the surface of the
Earth where observations are typically expressed in non-inertial coordinates
(latitude and longitude) that are fixed to the rotating Earth. In a coordinate
system ($x'_1, x'_2$) aligned with a tangential plane to the Earth, the
angular velocity vector $\vec{\omega}$ in the northern hemisphere has a
component  $\vec{\omega}_\perp$ that points away from the surface of the
earth.

\begin{figure*}[h]
\centerline{\includegraphics{coriolis2.eps}}
\end{figure*}

If a mass moves in the tangential plane with a velocity $\vec{v'}$, the 
projection of the Coriolis force $\vec{F}_C$ onto the tangential plane gives
rise to a force that pulls the mass to the right side (with respect to its
velocity, and seen when standing on the tangential plane). Consequently, air
masses that move into a low pressure area on the northern hemisphere form a
spiral that rotates counterclockwise around the low pressure area. However,
only the component $\vec{\omega}_\perp$ of the angular velocity that is
perpendicular to the tangential plane ($x'_1, x'_2$) contributes to an
in-plane projection of the Coriolis force  $\vec{F}_{C\,||}$. 

On the equator, $\vec{\omega}_\perp=0$ because the angular velocity vector
$\vec{\omega}$ is parallel to the surface. For velocities $\vec{v'}$
tangential to the surface, the Coriolis force has therefore no component
$\vec{F}_{C\,||}$ in the tangential plane.  
On the southern hemisphere, the vertical component of the angular velocity
vector points into the earth, leading to air moving into low pressure areas
in clockwise spiraling motion.


The Coriolis force is also the basis of recent solid-state inertial
sensors for rotation: in so-called {\em
  \href{https://en.wikipedia.org/wiki/Vibrating_structure_gyroscope}{vibrating
    structure gyroscopes}} or {\em
  \href{https://books.google.com.sg/books?id=WgFPvZyApd0C&redir_esc=y}{MEMS
    gyroscopes}}, an oscillating test mass feels the Coriolis force
perpendicular to its oscillating motion if there is a rotation component
perpendicular to the motion of the test mass. A similar effect seems to be
used by some two-winged insects, which sense forces on so-called {\em \href{https://en.wikipedia.org/wiki/Halteres}{halteres}}
vibrating with the wing frequency to determine their angular velocity and
presumably stabilize their orientation in space.

\clearpage
\section{Motion of rigid bodies}
In the last section, we encountered a few effects related with the
transformation between an inertial reference frame, and a rotating reference
frame. However, we did not ask about the dynamics of a rotational motion. In
this last section, we will address some of these aspects for so-called rigid
bodies, i.e., an ensemble of mass points in a particular shape with a fixed
relative position to each other.


\subsection{Orientation of a rigid body in space - Euler angles}
Several properties of rigid bodies are easy to describe in a coordinate system
that is fixed to the rigid body. Such a coordinate system may not be
an inertial reference frame, so we need to describe the relative
orientation of the body coordinate system $F'$ with respect to a fixed
inertial system $F$. In section \ref{ssec:coordinatetransformations} we saw
that this transformation can  be expressed by a proper rotation, so we
just need to parameterize this rotation.
A common way to do this is to compose the rotation
by three rotations around fixed axes such that a vector transforms as
\begin{equation}\label{eq:eulerangles}
  \vec{x'}=\tensor{R}_3(\psi)\cdot\tensor{R}_1(\theta)\cdot\tensor{R}_3(\phi)\cdot\vec{x}\,,
\end{equation}
with rotation matrices (see section \ref{ssec:coordinatetransformations})
\begin{equation}
\tensor{R}_3(\phi)=\left(\begin{array}{ccc}
      \cos\phi&-\sin\phi&0\\\sin\phi&\cos\phi&0\\0&0&1
      \end{array}\right)\quad{\rm and}\quad
\tensor{R}_1(\theta)=\left(
    \begin{array}{ccc}
      1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta
    \end{array}\right)\,.
\end{equation}
The angles $\psi, \theta, \phi$ in (\ref{eq:eulerangles}) are referred to as
{\em Euler angles}\footnote{again after
  \href{https://en.wikipedia.org/wiki/Leonhard_Euler}{Leonhard Euler},
  1707-1783}. The first rotation is around the $x_3$ axis, followed by a
rotation around the $x_1$ axis, followed again by a rotation around the $x_3$
axis. Such a combination of rotations
allows preparing any orientation of a rigid body in space. The figure below
shows the orientation of the coordinate axes in the intermediate and final
steps (for positive rotation angles $\phi, \theta $ and $\psi$):

\begin{figure*}[h]
\centerline{\includegraphics{eulerangles.eps}}
\end{figure*}
 
To work with the Euler angles in practice, the matrix multiplication
(\ref{eq:eulerangles}) has to be directly carried out\footnote{To worsen this
  misery, not all texts use the same convention on how to count the
  Euler angles. Check carefully if you ever need them, or avoid them
  altogether when you can.}:
\begin{eqnarray}
\tensor{R}&=&\tensor{R}_3(\psi)\cdot\tensor{R}_1(\theta)\cdot\tensor{R}_3(\phi)
\\
&=&\left(\begin{array}{ccc}
    \begin{array}{l}\cos\psi\cos\phi\\\,-\sin\psi\cos\theta\sin\phi\end{array}&
    \begin{array}{l}-\cos\psi\sin\phi\\\,-\sin\psi\cos\theta\cos\phi\end{array}&
    \sin\psi\sin\theta\\[1.5em]
    \begin{array}{l}\sin\psi\cos\phi\\\,+\cos\psi\cos\theta\sin\phi\end{array}&
    \begin{array}{l}-\sin\psi\sin\phi\\\,+\sin\psi\cos\theta\cos\phi\end{array}&
    -\cos\psi\sin\theta\\[1.5em]
    \sin\theta\sin\phi&\sin\theta\cos\phi&\cos\theta
  \end{array}\right)\,.\nonumber
\end{eqnarray}

\subsubsection{Euler angles and angular velocity}
In some situations, one can use the Euler angles as dynamic variables that
describe the dynamics of the orientation of a body in an inertial reference
frame. Then, the time derivatives of the Euler angles will become the
generalized velocities. The changing angles can be represented by vectors
$\dot{\vec{\phi}}$, $\dot{\vec{\theta}}$, and $\dot{\vec{\psi}}$
pointing in the direction of the rotation axis as discussed in section
\ref{ssec:smallrotations}:

\begin{figure*}[h]
\centerline{\includegraphics{eulerangles2.eps}}
\end{figure*}
Vector $\dot{\vec{\phi}}$ is aligned with the $x_3$ axis in the inertial
system $F$, vector $\dot{\vec{\psi}}$ points into the direction of the $x'_3$
axis in system $F'$ attached to the rigid body, and $\dot{\vec{\theta}}$
points into the direction of the {\em line of nodes} where the two circles in
the figure intersect. The instant values of $\dot{\vec{\phi}}$,
$\dot{\vec{\theta}}$, and $\dot{\vec{\psi}}$ can be summed up to an angular
velocity of the rigid body with respect to the inertial frame $F$. To be more
useful in describing the dynamics later, we will describe this vector in
system $F'$ attached to the body.

The simplest case is that of the angular velocity associated to a change of
Euler angle $\psi$, because it is already aligned with the basis vector 3 of
$F'$. therefore, the components of $\dot{\vec{\psi}}$ in $F'$ are given by:
\begin{equation}
\dot{\vec{\psi}}'_1=\dot{\vec{\psi}}'_2=0\,,\quad\dot{\vec{\psi}}'_3=\dot{\psi}\,, 
\end{equation}
where $\dot{\psi}$ is simply the change of Euler angle $\psi$ in time.
Vector $\dot{\vec{\theta}}$ is parallel to the line of nodes, and therefore
perpendicular to the $x'_3$ axis; its components in $F'$ are given by
\begin{equation}
\dot{\vec{\theta}}'_1=\dot{\theta}\cos\psi\,,\quad
\dot{\vec{\theta}}'_2=-\dot{\theta}\sin\psi\,,\quad
\dot{\vec{\theta}}'_3=0\,.
\end{equation}
Finally, the components of $\dot{\vec{\phi}}$ in $F'$ are given by
\begin{equation}
\dot{\vec{\phi}}'_1=\dot{\phi}\sin\theta\sin\psi\,,\quad
\dot{\vec{\phi}}'_2=\dot{\phi}\sin\theta\cos\psi\,,\quad
\dot{\vec{\phi}}'_3=\dot{\phi}\cos\theta\,.
\end{equation}
The total angular velocity vector $\vec{\omega}$ due to a change of Euler
angles in time is given by the sum of individual rotation vectors. In the
reference system $F'$ attached to the body, the components of $\vec{\omega}$
are therefore given by
\begin{equation}\label{eq:eulerangularvelocity}
\vec{\omega}\,=\,\dot{\vec{\phi}}+\dot{\vec{\theta}}+\dot{\vec{\psi}}
\,=\,\left(\begin{array}{lclcl}
\dot{\phi}\sin\theta\sin\psi&+&\dot{\theta}\cos\psi&&\\
\dot{\phi}\sin\theta\cos\psi&-&\dot{\theta}\sin\psi&&\\
\dot{\phi}\cos\theta&&&+&\dot{\psi}\\
\end{array}\right)\,.
\end{equation}

\subsubsection{Limitations of Euler angles to describe body orientations}
While every rotation of a rigid body with respect to an inertial system can be
described by a set of Euler angles, its usefulness for practical problems
is very limited. The problem is that not every rotation leads to a smooth
evolution of the Euler angles. For example, a body reference system $F'$ and
the inertial reference system $F$ are initially aligned, a rotation around
axis $x_2$ would require the angle $\phi$ to jump to $\pi/2$ and $\psi$ to
$-\pi/2$, while $\theta$ follows the rotation. This problem is referred to as
{\em \href{https://en.wikipedia.org/wiki/Gimbal_lock}{gimbal lock}}, and was a
problem in early inertial guidance systems, among them one that occurred during
the lunar landing mission of Apollo 11.

Therefore, other ways of representing the orientation of a body are often
chosen. One can use the rotation matrix $\tensor{R}$ directly, but would have
to store 9 entries. While this allows for fast vector transformations, this
method has the disadvantage that it is numerically intensive to correct for
numerical errors in matrices that are not exact rotations.

A method that requires four variables to store the orientation of a body, and
avoids the gimbal lock problem as well as the re-normalization problem involves
{\em
  \href{https://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation}{quaternions}}. 
They are an extension of the concept of complex numbers, with three roots of
-1 and allow for an efficient calculation of concatenated
rotations. Quaternions are beyond the scope of this course, but they form the
basis of most orientation descriptions in current applications in navigation,
robotics and computer graphics.

\clearpage
\subsection{Inertia of a rigid body}
A description of the dynamics of a rotating body requires
an expression for its kinetic energy. We consider the rigid body to be made up
by $N$ masses $m_\alpha$, with a fixed relative position with respect to each
other. The positions of the masses forming the body are described in a
coordinate system or reference frame $F'$ fixed to the body, because there,
the positions do not depend on the orientation of the body in space.
The position $\vec{r}_\alpha$ of each particle is described by a
center-of-mass position $\vec{R}$ of the body with respect to some inertial
reference frame $F$ with origin $O$, and a distance vector
$\vec{r'}_\alpha$ from the center-of-mass location:

\begin{figure*}[h]
\centerline{\includegraphics{rigidbody1.eps}}
\end{figure*}
\noindent By
definition, the velocity of the individual masses making up the rigid body
vanishes identically in the reference frame $F'$ moving with the body:
\begin{equation}
\vec{v'}_\alpha=\left.\frac{d\vec{r'}_\alpha}{dt}\right|_{F'}\equiv 0\,.
\end{equation}
Using (\ref{eq:timederivativebasevec}), the velocity of mass $m_\alpha$ in
inertial reference frame $F$ is given by
\begin{eqnarray}\label{eq:velocityinrigidbody}
\vec{v}_\alpha&=&\left.\frac{d\vec{r}_\alpha}{dt}\right|_F=
\frac{d\vec{R}}{dt}+\left.\frac{d\vec{r'}_\alpha}{dt}\right|_{F}
=\frac{d\vec{R}}{dt}+\underbrace{\left.\frac{d\vec{r'}_\alpha}{dt}\right|_{F'}}_{=0}+\vec{\omega}\times\vec{r'}_\alpha\nonumber\\
&=&\vec{V}+\vec{\omega}\times\vec{r'}_\alpha\,,
\end{eqnarray}
where $\vec{V}$ denotes the center-of-mass velocity of the body, and $\vec{\omega}$ is the instantaneous rotation vector of the moving
reference frame $F'$ with respect to $F$. 

\subsubsection{Kinetic energy of a rigid body}
With this, the kinetic energy of the body can be evaluated:
\begin{eqnarray}
T&=&\frac{1}{2}\sum\limits_\alpha m_\alpha \vec{v}_\alpha^2=
\frac{1}{2}\sum\limits_\alpha m_\alpha
\left(\vec{V}+\vec{\omega}\times\vec{r'}_\alpha\right)^2\nonumber\\
&=&\frac{1}{2}\sum\limits_\alpha m_\alpha\vec{V}^2\,+\,\sum\limits_\alpha m_\alpha
\vec{V}\cdot(\vec{\omega}\times\vec{R'}_\alpha)\,+\,\frac{1}{2}\sum\limits_\alpha
m_\alpha(\vec{\omega}\times\vec{r'}_\alpha)^2\\
&=&\frac{1}{2}\underbrace{\left(\sum\limits_\alpha m_\alpha\right)}_{=M}\vec{V}^2
\,+\,\vec{V}\cdot\Bigg[\vec{\omega}\times\underbrace{\sum\limits_\alpha m_\alpha\vec{r'}_\alpha}_{=M\vec{R'}}\Bigg]
\,+\,\frac{1}{2}\sum\limits_\alpha m_\alpha(\vec{\omega}\times\vec{r'}_\alpha)^2\,,
\nonumber
\end{eqnarray}
where $M$ is the total mass of the body. The second sum over
masses contains the center-of-mass position $\vec{R'}$
in the body coordinate system $F'$. If the reference system $F'$ is centered
in the center-of-mass position of the body then $\vec{R'}=0$, and the total
kinetic energy can be written as
\begin{equation}
T=\frac{1}{2}M\vec{V}^2+\frac{1}{2}\sum\limits_\alpha
m_\alpha(\vec{\omega}\times\vec{r'}_\alpha)^2=:T_{\rm trans}+T_{\rm rot}\,, 
\end{equation}
i.e., the total kinetic energy is a sum of a translation part that only
depends on the center-of-mass velocity $\vec{V}$ and the total mass $M$  of the
body, and a  {\em rotational energy} $T_{\rm rot}$ that is independent from
the center-of-mass motion. 

The modulus of the vector product in the rotational energy can be converted to
scalar products with the vector identity
\begin{equation}
(\vec{a}\times\vec{b})^2=(\vec{a}\times\vec{b})\cdot(\vec{a}\times\vec{b})=
(\vec{a}\cdot\vec{a})(\vec{b}\cdot\vec{b})-(\vec{a}\cdot\vec{b})^2\,,
\end{equation}
leading to the rotational energy
\begin{equation}
T_{\rm rot}=\frac{1}{2}\sum\limits_\alpha m_\alpha \left[\vec{\omega}^2 \vec{r'}_\alpha^2-(\vec{\omega}\cdot\vec{r'}_\alpha)^2\right]\,.
\end{equation}
We now evaluate the scalar products explicitly in Cartesian components:
\begin{eqnarray}\label{eq:totalrotationenergy0}
T_{\rm
  rot}&=&\frac{1}{2}\sum\limits_\alpha m_\alpha\left[\left(\sum\limits_{i=1}^{3}\omega_i^2\right)\left(\sum\limits_{k=1}^{3}{x'}_{\alpha,k}^2\right)-\left(\sum\limits_{i=1}^{3}\omega_i{x'}_{\alpha,i}\right)\left(\sum\limits_{j=1}^{3}\omega_j{x'}_{\alpha,j}\right)\right]\nonumber\\
&=&\sum\limits_\alpha\sum\limits_{i,j=1}^3
m_\alpha\left[\omega_i\omega_j\delta_{ij}\left(\sum\limits_{k=1}^{3}{x'}_{\alpha,k}^2\right)-\omega_i\omega_j{x'}_{\alpha,i}{x'}_{\alpha,j}\right]\\
&=&\frac{1}{2}\sum\limits_{i,j=1}^3\omega_i\omega_j\underbrace{\left[\sum\limits_\alpha
    m_\alpha\left(\delta_{ij}\left(\sum\limits_{k=1}^{3}{x'}_{\alpha,k}^2\right)-{x'}_{\alpha,i}{x'}_{\alpha,j}\right)\right]}_{=:I_{ij}}
=\frac{1}{2}\sum\limits_{i,j=1}^3\omega_i\omega_jI_{ij}\nonumber\,.
\end{eqnarray}
The total kinetic energy is therefore a bilinear function of the
vector $\vec{\omega}$, with coefficients $I_{ij}$ that only depend on the mass
distribution in the rigid body. The coefficients $I_{ij}$ can be written in
matrix, and the kinetic energy becomes a ``sandwich product'' of a matrix
between two vectors,
\begin{equation}\label{eq:totalrotationenergy}
T_{\rm
  rot}=\frac{1}{2}\,\vec{\omega}\cdot\tensor{I}\cdot\vec{\omega}\,,\quad{\rm
  with}\quad
\tensor{I}=\left(\begin{array}{ccc}
I_{11}&I_{12}&I_{13}\\I_{21}&I_{22}&I_{23}\\I_{31}&I_{32}&I_{33}\end{array}\right)\,.
\end{equation}
The object $\tensor{I}$ is referred to as the {\em inertia tensor} for the
rigid body. A tensor associated with a physical property has slightly richer
properties than a simple matrix in the sense that it has well-defined
transformation properties under similarity transformations. Before looking
into this, we review the properties of the matrix entries of $\tensor{I}$:
\begin{eqnarray}\label{eq:inertiatensor}
\tensor{I}=\left(\begin{array}{ccc}
\sum\limits_\alpha m_\alpha({r'}_\alpha^2-{x'}_{\alpha,1}^2)&
-\sum\limits_\alpha m_\alpha x'_{\alpha,1} x'_{\alpha,2}&
-\sum\limits_\alpha m_\alpha x'_{\alpha,1} x'_{\alpha,3}\\
-\sum\limits_\alpha m_\alpha x'_{\alpha,2} x'_{\alpha,1}&
\sum\limits_\alpha m_\alpha({r'}_\alpha^2-{x'}_{\alpha,2}^2)&
-\sum\limits_\alpha m_\alpha x'_{\alpha,2} x'_{\alpha,3}\\
-\sum\limits_\alpha m_\alpha x'_{\alpha,3} x'_{\alpha,1}&
-\sum\limits_\alpha m_\alpha x'_{\alpha,3} x'_{\alpha,2}&
\sum\limits_\alpha m_\alpha({r'}_\alpha^2-{x'}_{\alpha,3}^2)
\end{array}\right)\,.
\end{eqnarray}
The $x'_{\alpha,j}$ refer to the $j$-th component of
the position of mass element $\alpha$ with respect to the center-of-mass of
the body, and
${r'}^2_\alpha={x'}^2_{\alpha,1}+{x'}^2_{\alpha,2}+{x'}^2_{\alpha,3}$ is the
square of the distance of mass element $\alpha$ from the center-of-mass.
The tensor entries are symmetric under index exchange:
\begin{equation}\label{eq:tensorsymmetry1}
I_{ij}=I_{ji}\,,
\end{equation}
which reduces the independent tensor elements to three diagonal terms
$I_{11}$, $I_{22}$, and $I_{33}$ (referred to as {\em moments of inertia}),
and three independent off-diagonal elements $I_{12}=I_{21}$, $I_{23}=I_{32}$,
and $I_{13}=I_{31}$ (referred to as {\em products of inertia}). 

\subsubsection{Evaluation of the tensor elements}
As shown in the figure below for the example of axis $x_3$, the moments of
inertia (diagonal 
elements of $\tensor{I}$) in (\ref{eq:inertiatensor}) contain the difference
\begin{equation}
{r'}_\alpha^2-{x'}_{\alpha,3}^2={r'}_{\alpha,\perp}^2\,,
\end{equation}
where ${r'}_\perp$ is the distance to the rotation axis of mass
element $\alpha$. This reproduces the form for the moment of
inertia known from elementary physics courses.

\begin{figure*}[h]
\centerline{\includegraphics{inertialmoment.eps}}
\end{figure*}

To evaluate the inertia tensor elements for continuous solids, the sum over
all masses in (\ref{eq:inertiatensor}) is replaced by an integral over the
volume $V$ of the body, and the masses $m_\alpha$ by a (possibly
position-dependent) mass density $\rho(\vec{r'})$:
\begin{equation}
I_{ij}=\int\limits_Vd^3x'\rho(\vec{r'})\left[\delta_{ij}\vec{r'}^2-x'_ix'_j\right]\,,
\end{equation}
with volume element $d^3x'$ for the integration.

\subsubsection{Angular momentum of a rigid body}
From (\ref{eq:totalangular3}) in section
\ref{ssec:angularmomentum}, a rigid body has a total angular momentum
\begin{equation}
\vec{L}=\sum\limits_\alpha\vec{r}_\alpha\times\vec{p}_\alpha=\vec{R}\times\vec{P}+\sum\limits_\alpha\vec{r}'_\alpha\times\vec{p}'_\alpha\,,
\end{equation}
with center-of-mass position $\vec{R}$ and total linear momentum $\vec{P}$.
In the reference frame $F'$ fixed to the body (with an origin at its
center-of-mass), this simplifies to
\begin{equation}\label{eq:rigidtotalangular1}
\vec{L'}=\sum\limits_\alpha \vec{r'}_\alpha\times\vec{p'}_\alpha\,.
\end{equation}
The velocity $\vec{v'}_\alpha$ in $\vec{p'}_\alpha$  is 
given by (\ref{eq:velocityinrigidbody}) in an inertial reference frame
$F$. It is referenced with
respect to the center-of-mass of the body where $\vec{V'}=0$, so 
\begin{equation}
\vec{p'}_\alpha=m_\alpha \vec{v'}_\alpha=m_\alpha(\vec{\omega}\times\vec{r'}_\alpha)\,.
\end{equation}
With this, the total angular momentum (\ref{eq:rigidtotalangular1}) becomes
\begin{eqnarray}\label{eq:rigidtotalangular2}
\vec{L}'&=&\sum\limits_\alpha m_\alpha\vec{r'}_\alpha\times(\vec{\omega}\times\vec{r'}_\alpha)\nonumber\\
&=&\sum\limits_\alpha m_\alpha\left[{r'}^2_\alpha\vec{\omega}-\vec{r'}_\alpha(\vec{r'}_\alpha\cdot\vec{\omega})\right]\,,
\end{eqnarray}
where in the last step, we made use of the vector product identity
\begin{equation}
\vec{a}\times(\vec{b}\times\vec{a})=(\vec{a}\cdot\vec{a})\vec{b}-\vec{a}(\vec{a}\cdot\vec{b})\,.
\end{equation}
Expression (\ref{eq:rigidtotalangular2}) has two entries that are linear in
$\vec{\omega}$; to see this better, we look at the components of $\vec{L'}$:
\begin{eqnarray}
L'_i&=&\sum\limits_\alpha
m_\alpha\left[{r'}_\alpha^2\omega_i-{x'}_{\alpha,i}\sum\limits_j {x'}_{\alpha,j}\omega_j\right]\nonumber\\
&=&\sum\limits_\alpha m_\alpha\sum\limits_j\left[\omega_j\delta_{ij}\sum\limits_k{x'}_{\alpha,k}^2-{x'}_{\alpha,i}{x'}_{\alpha,j}\omega_j\right]\nonumber\\
&=&\sum\limits_j\underbrace{\left[\sum\limits_\alpha m_\alpha\left(\delta_{ij}\sum\limits_k{x'}_{\alpha,k}^2-{x'}_{\alpha,i}{x'}_{\alpha,j}\right)\right]}_{=I_{ij}}=\sum\limits_jI_{ij}\omega_j
\end{eqnarray}
with the same tensor components $I_{ij}$ as in (\ref{eq:totalrotationenergy0}). 
The last expression is the result of a multiplication of vector $\vec{\omega}$
from the right side to tensor $\tensor{I}$, so the total angular momentum can
also be written as:
\begin{equation}\label{eq:rigidangular3}
\vec{L'}=\tensor{I}\cdot\vec{\omega}\,,
\end{equation}
which is the rotation analog to the expression of the linear momentum
$\vec{p}=m\vec{v}$. Inertia tensor $\tensor{I}$ therefore takes the
role of the inertial object property for rotations, but it is not a simple
scalar like the mass $m$. The total rotational
energy (\ref{eq:totalrotationenergy}) can be expressed as a scalar product
using the angular momentum:
\begin{equation}
T_{\rm rot}\,=\,\frac{1}{2}\,\vec{\omega}\cdot(\tensor{I}\cdot\vec{\omega})
\,=\,\frac{1}{2}\,\vec{\omega}\cdot\vec{L'}\,.
\end{equation}

\subsubsection{Transformation properties of the inertia tensor}
In section \ref{ssec:coordinatetransformations}, we briefly mentioned that
vectors and tensors as physical properties can be identified by their
transformation properties under similarity transformations $\tensor{M}$. 
To see this, we take (\ref{eq:rigidangular3}) and change the notation
slightly, since the prime in $\vec{L'}$ just indicates that the angular
momentum was to be taken with respect to a coordinate system originating in
the COM of the rigid body. We now consider two versions of
(\ref{eq:rigidangular3}) in two coordinate systems $F$ and $F'$ that both
originate in the  COM of the rigid body:
\begin{equation}\label{eq:tensortrans}
\vec{L}=\tensor{I}\cdot\vec{\omega}\quad{\rm and}\quad
\vec{L'}=\tensor{I'}\cdot\vec{\omega'}\,.
\end{equation}
Both $\vec{L}$ and $\vec{\omega}$ are vectors, so they obey the transformation
rules 
\begin{equation}
\vec{L'}=\tensor{M}\cdot\vec{L}\quad{\rm and}\quad
\vec{\omega'}=\tensor{M}\cdot\vec{\omega}
\end{equation}
with the same matrix $\tensor{M}$. Using this in (\ref{eq:tensortrans}) leads
to
\begin{equation}
\tensor{M}\cdot\tensor{I}\cdot\vec{\omega}=\tensor{M}\cdot\vec{L}=\vec{L'}=\tensor{I'}\cdot\vec{\omega'}=\tensor{I'}\cdot\tensor{M}\cdot\vec{\omega}\,.
\end{equation}
This equality holds for all vectors $\vec{\omega}$, leading to the matrix
identity
\begin{equation}
\tensor{M}\cdot\tensor{I}=\tensor{I'}\cdot\tensor{M}
\end{equation}
or
\begin{equation}\label{eq:tensortransform1}
\tensor{I'}=\tensor{M}\cdot\tensor{I}\cdot\tensor{M}^{-1}
\end{equation}
which is the matrix version of the componentwise tensor transformation rule
(\ref{eq:tensortransform}). This can be seen using
\begin{equation}
(\tensor{M}^{-1})_{ij}=(\tensor{M}^T)_{ij}=(\tensor{M})_{ji}
\end{equation}
for similarity transformations $\tensor{M}$. Then,
\begin{eqnarray}
I'_{ij}=(\tensor{I'})_{ij}&=&\left(\tensor{M}\cdot\tensor{I}\cdot\tensor{M}^{-1}\right)_{ij}\nonumber\\
&=&\sum\limits_{kl}(\tensor{M})_{ik}(\tensor{I})_{kl}(\tensor{M}^{-1})_{lj}\nonumber\\
&=&\sum\limits_{kl}m_{ik}I_{kl}m_{jl}\,,
\end{eqnarray}
which is exactly the required transformation rule (\ref{eq:tensortransform})
for a tensor of rank 2. So the inertia tensor $\tensor{I}$ is a property of
the rigid body, independent of the chosen coordinate system in which it is
described.

\subsubsection{Principal rotation axes of a rigid body}
Relation (\ref{eq:rigidangular3}) states that the inertia tensor
$\tensor{I}$ linearly transforms the vector $\vec{\omega}$ into another
vector $\vec{L}$; if $\vec{\omega}$ and $\vec{L}$ are parallel, they are
eigenvectors of $\tensor{I}$. Each tensor $\tensor{I}$ has a set of three
orthogonal eigenvectors $\vec{i}_1, \vec{i}_2, \vec{i}_3$. In a coordinate
system where base vectors  $\vec{e}_i$ are parallel to the
eigenvectors $\vec{i}_i$, the inertia tensor takes a simple diagonal form:
\begin{equation}
\tensor{I}=\left(\begin{array}{ccc}
I_{11}&0&0\\0&I_{22}&0\\0&0&I_{33}\end{array}\right)=:
\left(\begin{array}{ccc}
I_1&0&0\\0&I_2&0\\0&0&I_3\end{array}\right)
\end{equation}
The directions $\vec{e}_i$ are called {\em principal rotation axes}, and for a
rotation around them, the angular momentum $\vec{L}$ is parallel to
$\vec{\omega}$. The evaluation of the inertial tensor is particularly simple
if the tensor $\tensor{I}$ is evaluated in coordinates aligned with the
principal rotation axes, as only three tensor elements need to be evaluated.

\subsubsection{Relation between body symmetries and the inertia tensor}\label{ssec:bodysummetries}
Evaluation of the inertial tensor is simple in the principal rotation axes,
but it is not always obvious how these axes are oriented. Often the
symmetry of the body gives an indication where they are. At the very least,
symmetries can help to reduce the calculation effort. To see how this works,
we first consider a body with a mirror symmetry. The example below has a
mirror symmetry with respect to plane $M_1$ perpendicular to the $x_1$ axis:

\begin{figure*}[h]
\centerline{\includegraphics{symmetry1.eps}}
\end{figure*}

The inertia tensor as a body property should therefore also be invariant under
a mirror transformation, represented by a matrix $\tensor{M}_1$:
\begin{equation}
\tensor{I}=\tensor{I'}=\tensor{M}_1\cdot\tensor{I}\cdot\tensor{M}_1^{-1}\quad{\rm
  with}\quad\tensor{M}_1=\tensor{M}_1^{-1}
=\left(\begin{array}{ccc}-1&0&0\\0&1&0\\0&0&1\end{array}\right)
\end{equation}
The symmetry requirement is equivalent to
\begin{equation}
\tensor{I}\cdot\tensor{M}_1=\tensor{M}_1\cdot\tensor{I}\,,
\end{equation}
which we now explicitly check by calculating the matrix products:
\begin{equation}
\tensor{I}\cdot\tensor{M}=\left(\begin{array}{ccc}
-I_{11}&I_{12}&I_{13}\\-I_{21}&I_{22}&I_{23}\\-I_{31}&I_{32}&I_{33}\end{array}\right)
\quad{\rm and}\quad
\tensor{M}\cdot\tensor{I}=\left(\begin{array}{ccc}
-I_{11}&-I_{12}&-I_{13}\\I_{21}&I_{22}&I_{23}\\I_{31}&I_{32}&I_{33}\end{array}\right)\,.
\end{equation}
Since both matrices need to be the same, we can immediately see by comparison
that $I_{12}=I_{21}=I_{13}=I_{31}=0$.  The inertia tensor of a body with mirror
symmetry in $x_1$ axis has therefore the form
\begin{equation}
\tensor{I}=\left(\begin{array}{ccc}\cdot&0&0\\0&\cdot&\cdot\\0&\cdot&\cdot
\end{array}\right)\,,
\end{equation}
where the dots indicate positions in the tensor that may not vanish. This
matrix is block diagonal, so we already know that a vector parallel to the
$x_1$ axis is an eigenvector of $\tensor{I}$, i.e., $x_1$ is a principal
rotation axis of the rigid body. Similarly,
one can show that the mirror symmetry in $x_3$ direction implies a form of the
inertia tensor of
\begin{equation}
\tensor{I}=\left(\begin{array}{ccc}\cdot&\cdot&0\\\cdot&\cdot&0\\0&0&\cdot
\end{array}\right)\,,
\end{equation}
so the $x_3$ axis is also a principal rotation axis. Therefore, the $x_2$ axis
is also a principal rotation axis, and tensor $\tensor{I}$ is diagonal in the
shown coordinates. This method of identifying vanishing tensor entries due to
symmetries requires very little effort, and simplifies the determination of
tensor entries a lot.

The shown example has another symmetry: it is invariant under rotations around
the $x_3$ axis by 120$^\circ$. We consider the implications of this symmetry
by looking at a rotation invariance under any rotation around the $x_3$ axis:
\begin{equation}
\tensor{R}_3(\alpha)\cdot\tensor{I}=\tensor{I}\cdot\tensor{R}_3(\alpha)\quad{\rm
  with}\quad
\tensor{R}_3(\alpha)=\left(\begin{array}{ccc}
c&-s&0\\s&c&0\\0&0&1\end{array}\right)\,,
\end{equation}
where  $c=\cos\alpha$ and $s=\sin\alpha$. Carrying out the
two multiplications leads to
\begin{eqnarray}\label{eq:rotatedinertias}
\tensor{R}_3(\alpha)\cdot\tensor{I}&=&\left(\begin{array}{ccc}
cI_{11}-sI_{21}&cI_{12}-sI_{22}&cI_{13}-sI_{23}\\
sI_{11}+cI_{21}&sI_{12}+cI_{22}&sI_{13}+cI_{23}\\
I_{31}&I_{32}&I_{33}\end{array}\right)\,,\nonumber\\
\tensor{I}\cdot\tensor{R}_3(\alpha)&=&\left(\begin{array}{ccc}
cI_{11}+sI_{12}&-sI_{11}+cI_{12}&I_{13}\\
cI_{21}+sI_{22}&-sI_{21}+cI_{22}&I_{23}\\
cI_{31}+sI_{32}&-sI_{31}+cI_{32}&I_{33}\end{array}\right)\,,
\end{eqnarray}
which we now explore more carefully. By comparing the $(1,1)$ entries in both
matrices, we find
\begin{equation}
-sI_{21}=sI_{12}=sI_{21}
\end{equation}
because of the symmetry $I_{ij}=I_{ji}$ from (\ref{eq:tensorsymmetry1}). For
rotation angles $\alpha\neq0,\pi,\dots$ where $s\neq0$, this requires that 
\begin{equation}
I_{12}=I_{21}=0\,.
\end{equation}
This, by comparing the $(1,2)$ components in (\ref{eq:rotatedinertias}),
requires also
\begin{equation}
I_{11}=I_{22}\,.
\end{equation}
Next, we compare the $(3,1)$ entries in the products
(\ref{eq:rotatedinertias}):
\begin{equation}
I_{31}=cI_{31}+sI_{32}\quad\rightarrow\quad\frac{I_{31}}{I_{32}}=\frac{s}{1-c}
\end{equation}
for a non-vanishing $I_{32}$. The same procedure for the $(3,2)$ entries gives
\begin{equation}
I_{32}=-sI_{31}+cI_{32}\quad\rightarrow\quad\frac{I_{31}}{I_{32}}=\frac{1-c}{-s}\,.
\end{equation}
Both ratios have different signs, which only can be fulfilled if
\begin{equation}
I_{31}=I_{32}=I_{13}=I_{23}=0\,.
\end{equation}
With these requirements, a tensor $\tensor{I}$  with a rotational symmetry
other than 180$^\circ$ in $x_3$ direction has the simple form
\begin{equation}\label{eq:symmetrictopinertia}
\tensor{I}=\left(\begin{array}{ccc}I_{11}&0&0\\0&I_{11}&0\\0&0&I_{33}\end{array}\right)\,,
\end{equation}
i.e., the axis of rotational symmetry is a principal rotation axis, and any
axis in the $x_1,x_2$ plane is a principal rotation axis as well, with
degenerate moments of inertia. A body with such properties is referred to as a
{\em symmetric top}.

The exploration of symmetries to determine the number of independent entries
in tensors goes far beyond the application to the tensor of inertia: In many
areas of physics, symmetries e.g. in materials (due to their crystalline
structure) imply if a physical property signified by a tensor takes a
particular form, or is even present.

\subsubsection{Inertia tensor for displaced rotation axes}
In the derivation of the inertia tensor components, we assumed that the mass
distribution is described in a coordinate system centered in the
center-of-mass of the rigid body.  However, often a rotation takes place around
an axes not containing the center-of-mass, so it is interesting to calculate
the inertia tensor elements with respect to a point $Q$ that is displaced from
the center-of-mass by a vector $-\vec{a}$:

\begin{figure*}[h]
\centerline{\includegraphics{rigidbody2.eps}}
\end{figure*}
To find the inertia tensor elements $J_{ij}$ with respect to the new center
$Q$, we recall expression (\ref{eq:totalrotationenergy0}) for the tensor
elements in the center-of-mass system,
\begin{equation}
I_{ij}=\sum\limits_\alpha m_\alpha\left[\delta_{ij}\left(\sum\limits_{k=1}^{3}{x'}_{\alpha,k}^2\right)-{x'}_{\alpha,i}{x'}_{\alpha,j}\right]\,,
\end{equation}
and evaluate the expression for displaced positions
$\vec{r}_\alpha=\vec{r'}_\alpha+\vec{a}$ for the masses:
\begin{eqnarray}
J_{ij}&=&\sum\limits_\alpha m_\alpha\left[\delta_{ij}\left(\sum\limits_{k=1}^{3}({x'}_{\alpha,k}+a_k)^2\right)-({x'}_{\alpha,i}+a_i)({x'}_{\alpha,j}+a_j)\right]\nonumber\\
&=&\underbrace{\sum\limits_\alpha
m_\alpha\left[\delta_{ij}\left(\sum\limits_{k=1}^{3}{x'}_{\alpha,k}^2\right)-{x'}_{\alpha,i}\,{x'}_{\alpha,j}\right]}_{=I_{ij}}\nonumber\\
&&+\sum\limits_\alpha m_\alpha \left[\delta_{ij}\left(\sum\limits_{k=1}^{3}(a_k^2+2{x'}_{\alpha,k}a_k)\right)-(a_i{x'}_{\alpha,j}+a_j{x'}_{\alpha,i}+a_ia_j)\right]
\end{eqnarray}
In the second line, all terms that contain a single $x'_\alpha$ component
appear in a sum over all $\alpha$ together with mass $m_\alpha$. Since the
$x'_{\alpha,k}$ are positions relative to the center-of-mass,
\begin{equation}
\sum\limits_\alpha m_\alpha x'_{\alpha,k}=0
\end{equation}
by definition. This simplifies the expression of the new tensor elements:
\begin{eqnarray}
J_{ij}&=&I_{ij}+\sum\limits_\alpha
m_\alpha\left[\delta_{ij}\left(\sum\limits_{k=1}^{3}a_k^2\right)-a_ia_j\right]\nonumber\\
&=&I_{ij}+M[\delta_{ij}\vec{a}^2-a_ia_j]\,,
\end{eqnarray}
where $M$ is the total mass of the rigid body. Thus, the inertia tensor
$\tensor{J}$  with
respect to a different rotation center $Q$ than the center-of-mass is given by
the sum of the inertia 
tensor $\tensor{I}$ of the body with respect to its center-of-mass and an
inertia tensor by a single mass $M$, displaced by a vector $\vec{a}$ from the
rotation center. This is the so-called {\em parallel axis--} or {\em
  Huygens-Steiner} theorem\footnote{after
    \href{https://en.wikipedia.org/wiki/Christiaan_Huygens}{Christiaan
      Huygens}, 1629-1695 and
    \href{https://en.wikipedia.org/wiki/Jakob_Steiner}{Jakob Steiner},
    1796-1863}.

\subsection{Equation of motion of a rotating rigid body}
To describe the dynamics of a rotating body, we first choose a coordinate
system $F'$ fixed to the body that is aligned with the principal rotation
axes. Then, the inertia tensor is diagonal, and the rotational energy
(\ref{eq:totalrotationenergy}) and angular momentum (\ref{eq:rigidangular3})
are  given by
\begin{equation}
T_{\rm rot}=\frac{1}{2}\sum\limits_{i=1}^3I_i\omega_i^2\quad{\rm and}\quad
L_i=I_i\omega_i\,.
\end{equation}
The influence of external forces to the body is captured by a torque $\vec{N}$
acting on the body; we recall (\ref{eq:changeofangularmomentum}):
\begin{equation}
\frac{d\vec{L}}{dt}=\vec{N}\,.
\end{equation}
This relation was derived in an inertial system, but it is most convenient to
describe the dynamics of rotation in the coordinate system $F'$ attached to
the body. Therefore, we use the transformation rule
(\ref{eq:timederivativerot}) for time derivative of vectors between reference
systems:
\begin{equation}
\vec{N}=\left.\frac{d\vec{L}}{dt}\right|_F=\left.\frac{d\vec{L}}{dt}\right|_{F'}+\vec{\omega}\times\vec{L}\,.
\end{equation}
We now evaluate this equation for one component in the reference system $F'$
attached to the body:
\begin{eqnarray}
\left.\frac{dL_1}{dt}\right|_F=N_1&=&\left.\frac{dL_1}{dt}\right|_{F'}+\omega_2L_3-\omega_3L_2\nonumber\\
&=&\left.\frac{d(I_1\omega_1)}{dt}\right|_{F'}+\omega_2I_3\omega_3-\omega_3I_2\omega_2\nonumber\\
&=&I_1\dot{\omega}_1+\omega_3\omega_2(I_3-I_2)\,.
\end{eqnarray}
The equations for the other components can be obtained by cyclic permutation
of the indices:
\begin{eqnarray}
N_2&=&I_2\dot{\omega}_2+\omega_1\omega_3(I_1-I_3)\quad{\rm and}\nonumber\\
N_3&=&I_3\dot{\omega}_3+\omega_2\omega_1(I_2-I_1)\,.
\end{eqnarray}
These equation can be summarized in a compact notation with the totally
antisymmetric symbol $\epsilon_{ijk}$:
\begin{equation}\label{eq:eulerequationsrigid}
(I_i-I_j)\omega_i\omega_j-\sum\limits_k\epsilon_{ijk}(I_k\dot{\omega}_k-N_k)=0\,.
\end{equation}
These equations are referred to as {\em Euler's\footnote{same \href{https://en.wikipedia.org/wiki/Leonhard_Euler}{Leonhard Euler}
    as before} equation of motion} for the rigid body. They look relatively
innocent, but what makes them hard to solve in practice is that the
torque $\vec{N}$ must also be expressed in the reference system $F'$ attached to
the rotating body, which may require a knowledge of the instantaneous
orientation in an external inertial reference frame.
\subsection{Force-free rotation of a symmetric top}\label{ssec:forcefreetop}
The solution of Euler's equation of motion (\ref{eq:eulerequationsrigid})
becomes relatively simple if there is no torque acting on the rotating body,
and it is freely evolving. We start with the motion of a symmetric top with
moments of inertia $I_1=I_2\neq I_3$. The equations of motion take the form
\begin{eqnarray}
(I_1-I_3)\omega_2\omega_3-I_1\dot{\omega}_1&=&0\nonumber\\
(I_3-I_1)\omega_3\omega_1-I_1\dot{\omega}_2&=&0\nonumber\\
I_3\dot{\omega}_3&=&0\,.
\end{eqnarray}
From the last equation, we find $\dot{\omega}_3=0$ or $\omega_3=const.$, so
the first two equations can be re-arranged:
\begin{equation}
\begin{array}{lcrcr}\label{eq:coupledtop}
\dot{\omega}_1&=&-\left(\frac{I_3-I_1}{I_3}\omega_3\right)\omega_2&=&-\Omega\omega_2\\[1em]
\dot{\omega}_2&=&\left(\frac{I_3-I_1}{I_3}\omega_3\right)\omega_1&=&\Omega\omega_1\end{array}
\end{equation}
with a constant
\begin{equation}\label{eq:precessionfreq1}
\Omega:=\frac{I_3-I_1}{I_1}\omega_3\,.\end{equation}
This coupled set of equations can be easily solved by introducing a variable
$\eta:=\omega_1+i\omega_2$, and adding the two equations (\ref{eq:coupledtop})
accordingly:
\begin{eqnarray}
\dot{\omega}_1+i\dot{\omega}_2=-i\Omega(\omega_1+i\omega_2)&=&0\quad{\rm
  or}\nonumber\\
\dot{\eta}-i\Omega\eta=0\,.
\end{eqnarray}
This simple linear equation of motion has a solution
\begin{equation}\label{eq:freespin1}
\eta(t)=Ae^{i\Omega t}=A\cos\Omega t\,+\,iA\sin\Omega t\,,
\end{equation}
or, for the two components of the rotation vector,
\begin{eqnarray}
\omega_1(t)&=&A\cos\Omega t\nonumber\\
\omega_2(t)&=&A\sin\Omega t\nonumber\\
\omega_3(t)&=&const.
\end{eqnarray}
We could have introduced a phase shift in the solution (\ref{eq:freespin1}) to
meet any initial condition, but that would not have substantially altered the
solution for $\vec{\omega}(t)$. The angular velocity vector $\vec{\omega}$
(with modulus $|\vec{\omega}|=\sqrt{\omega_3^2+A^2}$ is {\em precessing}
around the coordinate axis $x'_3$ in the rotating reference frame $F'$ with a
precession angular frequency $\Omega$ from
(\ref{eq:precessionfreq1}). Depending on the relative magnitude of $I_1$ and
$I_3$, the precession vector $\vec{\Omega}$ (indicating the precession sense of
$\vec{\omega}$) is either parallel or antiparallel to $\vec{e'}_3$. 

\begin{figure*}[h]
\centerline{\includegraphics[width=10cm]{precession1.eps}}
\end{figure*}
This still describes the dynamics of $\vec{\omega}$ in the reference frame of
the body. To transform this back to an inertial reference
frame of an observer, we note that the force-free condition (and the fact that
we have assumed no dissipation) requires that the rotational energy is
constant:
\begin{equation}\label{eq:precessionenergy}
T_{\rm rot}=\frac{1}{2}\vec{\omega}\cdot\vec{L}=const.
\end{equation}
Furthermore, the total angular momentum of the system is conserved in an
inertial reference frame $F$, i.e., the
vector $\vec{L}$ has both a constant modulus and direction. Energy
conservation in the form of (\ref{eq:precessionenergy}) requires
that the projection of $\vec{\omega}$ onto $\vec{L}$ and therefore the angle
between them is constant as well. To understand the relative orientation of
vectors $\vec{L}$, $\vec{\omega}$, and the principal axis of rotation
$\vec{e'}_3$ (often referred to as the {\em figure axis} of the symmetric top
because of the symmetry relation discussed earlier), we consider the vector
product
\begin{equation}
\vec{\omega}\times\vec{e'}_3=\omega_2\vec{e'}_1-\omega_2\vec{e'}_2
\end{equation}
with the components $\omega_i$ in the rotating reference frame. This vector is
perpendicular both to $\vec{e'}_3$ and $\vec{\omega}$. The scalar product
\begin{equation}
\vec{L}\cdot(\vec{\omega}\times\vec{e'}_3)=I_1\omega_1\omega_2-I_2\omega_2\omega_1=0
\end{equation}
vanishes because $I_1=I_2$ in the symmetric top. Therefore, $\vec{L}$ is
perpendicular to $\vec{\omega}\times\vec{e'}_3$, which implies that  $\vec{L}$, $\vec{\omega}$ and $\vec{e'}_3$ are all in the same plane:

\begin{figure*}[h]
\centerline{\includegraphics[width=10cm]{precession2.eps}}
\end{figure*}
Depending on the ratio between $I_1$ and $I_3$, the orientation of the three
vectors is as shown in the figure; here, $\vec{\Omega}$ denotes the direction
of $\vec{\omega}$ precessing around the figure axis $\vec{e'}_3$. In both
cases, the angle between the figure axis and the angular momentum is constant,
and the figure axis precesses around $\vec{L}$ fixed in space.

\subsection{Rotation of a heavy symmetric top}
As an example of a more complex rotary motion, we consider the movement of a
heavy symmetric top that is spinning around its figure axis, touching the
ground at a single point $O$; the distance between the touching point $O$ and
the center-of-mass of the top $C$ should be $l$:
\begin{figure*}[h]
\centerline{\includegraphics{heavytop.eps}}
\end{figure*}

In section \ref{ssec:bodysummetries} we saw that a rotationally symmetric top
with a body coordinate system $F'$ chosen to include the figure axis has a
diagonal inertia tensor $\tensor{I}$ with two possibly different entries $I_1$
and $I_3$.

The challenging part of this problem is the gravitational acceleration leading
to a non-vanishing torque $\vec{N}$. This torque is well-defined in the
inertial frame, but needs to be transformed into the body coordinate system
$F'$ to use the equations of motion (\ref{eq:eulerequationsrigid}) to describe
the dynamics of the rotation vector. Furthermore, the rotation vector
$\vec{\omega}$ would need to be integrated to obtain the orientation --- a task
that can not be accomplished as easily as for the force-free top.

This problem can be tackled if the Euler angles $\phi$, $\theta$ and $\psi$
describing the orientation of the top with respect to the inertial reference
frame are taken as generalized coordinates for the problem. The equations of
motion are then obtained with the standard Euler-Lagrange mechanism
(\ref{eq:generaleulerlagrange}). For this, an expression for the kinetic
energy $T$ is needed; for the symmetric top, it is according to
(\ref{eq:totalrotationenergy}) and  (\ref{eq:symmetrictopinertia}) given by
\begin{equation}
T_{\rm rot}=\frac{1}{2}\sum\limits_iI_i\omega'_i=\frac{1}{2}\left[I_1({\omega'}^2_1+{\omega'}^2_2)+I_3{\omega'}^2_3\right]\,.
\end{equation}
Note that in order to have only a rotational part of the kinetic energy, the
inertia tensor elements need to be evaluated with respect to the origin $O$ of
$F$ and $F'$, and $O$ is {\em not} the center-of-mass of the top. This is not a
problem, because the figure is still rotationally 
symmetric with respect to $x'_3$, hence $\tensor{I}$ is diagonal and has only
two different entries $I_1$, $I_3$. Using the expression
(\ref{eq:eulerangularvelocity}) for the angular velocity $\vec{\omega}$ in the
body coordinate system, one obtains a total kinetic energy as a function of the
generalized coordinates and velocities:
\begin{equation}
T=\frac{1}{2}I_1\left(\dot{\phi}^2\sin^2\theta+\dot{\theta}^2\right)+\frac{1}{2}I_3\left(\dot{\phi}\cos\theta+\dot{\psi}\right)^2\,.
\end{equation}
The potential energy is simply given by the total mass and the center-of-mass
position,
\begin{equation}
U=Mgl\cos\theta\,,
\end{equation}
leading to the Lagrange function
\begin{equation}
L=T-U=\frac{1}{2}I_1\left(\dot{\phi}^2\sin^2\theta+\dot{\theta}^2\right)+\frac{1}{2}I_3\left(\dot{\phi}\cos\theta+\dot{\psi}\right)^2-Mgl\cos\theta\,.
\end{equation}
Immediately, one can identify $\phi$ and $\psi$ as cyclic coordinates, leading
to two constant corresponding generalized momenta 
\begin{eqnarray}\label{eq:otherprecession1}
p_\phi=\frac{\partial L}{\partial \dot{\phi}}&=&
\left(I_1\sin^2\theta+I_3\cos^2\theta\right)\dot{\phi}+I_3\cos\theta\dot{\psi}
=const.\,,\\\label{eq:otherprecession2}
p_\psi=\frac{\partial L}{\partial \dot{\psi}}&=&
I_3\left(\dot{\psi}+\dot{\phi}\cos\theta\right)=const.
\end{eqnarray}
These quantities are angular momenta, and more specifically, projections of
the total angular momentum $\vec{L}$ on the $x_3$ axis and $x'_3$ axis,
respectively. They can be used to obtain simple differential equations for the
Euler angles $\psi$ and $\phi$:
\begin{eqnarray}\label{eq:topmotion1}
\dot{\phi}&=&\frac{p_\phi-p_\psi\cos\theta}{I_1\sin^2\theta}\,,\\
\label{eq:topmotion2}
\dot{\psi}&=&\frac{p_\psi}{I_3}-\frac{(p_\phi-p_\psi\cos\theta)\cos\theta}{I_1\sin^2\theta}\,.
\end{eqnarray}
To solve the remaining problem, i.e., the time evolution of $\theta$, we follow
an approach of an {\em effective potential} similar to the central force
problem in section 
\ref{ssec:effectivepotential}. The system is
conservative, so the total energy $E$ should be conserved:
\begin{equation}
E=\frac{1}{2}I_1\left(\dot{\phi}^2\sin^2\theta+\dot{\theta}^2\right)+\underbrace{\frac{1}{2}I_3{\omega'}_3^2}_{=p_\psi^2/2I_3}+Mgl\cos\theta=const.
\end{equation}
By subtracting the (constant) kinetic energy term due to rotation
around the figure axis from $E$ one finds the relation
\begin{eqnarray}\label{eq:topeffectivepotential}
E-\frac{p_\psi^2}{2I_3}=:E'&=&\frac{1}{2}I_1\dot{\theta}^2+\frac{(p_\phi-p_\psi\cos\theta)^2}{2I_1\sin^2\theta}+Mgl\cos\theta\nonumber\\
&=:&\frac{1}{2}I_1\dot{\theta}^2+V_{\rm eff}(\theta)\,,
\end{eqnarray}
introducing an effective potential $V_{\rm eff}(\theta)$ that only depends on
the angular momentum constants $p_\psi$, $p_\phi$ and the coordinate $\theta$.

Formally, one could integrate (\ref{eq:topeffectivepotential}), and analogous
to (\ref{eq:centralprobperiode}) in the central force problem obtain a function 
\begin{equation} 
t(\theta)=\int\limits_{\theta_0}^\theta\frac{d\theta'}{\sqrt{\frac{2}{I_1}(E'-V_{\rm eff}(\theta'))}}\,,
\end{equation}
invert the result to obtain $\theta(t)$, and use (\ref{eq:topmotion1}/\ref{eq:topmotion2}) to
obtain the solutions for the other two coordinates $\phi(t)$,
$\psi(t)$. However, this can be reasonably done only numerically.

The type of motion can be characterized in a similar way as for the central
force problem by looking at $V_{\rm eff}(\theta)$:

\begin{figure*}[h]
\centerline{\includegraphics{vefftop.eps}}
\end{figure*}
For $p_\phi\neq p_\psi$, the effective potential limits the angle $\theta$ to
the interval between 0 and $\pi$ due to the $\sin^2\theta$ term in the
denominator of the effective potential in
(\ref{eq:topeffectivepotential}). The effective potential has a minimum at
$\theta_0$, which can lead to a precession of the top with a constant angle
$\theta_0$ if the total excess energy $E'$ is minimal. Otherwise, $\theta$
will oscillate between two extremal angles $\theta_1$ and $\theta_2$. This
oscillatory motion is referred to as {\em nutation}.  

\subsubsection{Precession of the heavy top without nutation}
The minimum in the effective potential that leads to a motion with a constant
angle $\theta_0$ can be obtained by differentiating the effective
potential. With the abbreviation
\begin{equation}
\beta:=p_\phi-p_\psi\cos\theta\,, 
\end{equation}
that also appears in (\ref{eq:otherprecession1}/\ref{eq:otherprecession2}) for
the other two Euler angles, the condition for a minimum of $v_{\rm eff}$ is
\begin{eqnarray}
0\,\stackrel{!}{=}\,
\left.\frac{\partial V_{\rm eff}}{d\theta}\right|_{\theta=\theta_0}&=&
\frac{\partial}{\partial\theta}\left[\frac{\beta^2}{2I_1}+Mgl\cos\theta\right]_{\theta=\theta_0}\nonumber\\
&=&
\frac{2\beta
  p_\psi\sin\theta_0\sin^2\theta_0-\beta^22\sin\theta_0\cos\theta_0}{2I_1\sin^4\theta_0}-Mgl\sin\theta_0\nonumber\\
&=&
\frac{\beta p_\psi\sin^2\theta_0-\beta^2\cos\theta_0-MglI_1\sin^4\theta_0}{I_1\sin^3\theta_0}\,,
\end{eqnarray}
or
\begin{equation}
\beta p_\psi\sin^2\theta_0-\beta^2\cos\theta_0-MglI_1\sin^4\theta_0=0\,.
\end{equation}
This equation has two solutions for $\beta$:
\begin{equation}\label{eq:precessionbeta}
\beta_\pm=\frac{p_\psi\sin^2\theta_0}{2\cos\theta_0}\left[1\pm\sqrt{1-\frac{4MglI_1\cos\theta_0}{p_\psi^2}}\,\right]\,.
\end{equation}
For the solutions to be real, the argument of the root must be
positive. Assuming that $\theta_0<\pi/2$ (i.e., the top is above the surface),
this requires
\begin{equation}
4MglI_1\cos\theta_0<p_\psi^2=I_3^2{\omega'}_3^2\,.
\end{equation}
This condition imposes a minimum for the necessary angular momentum projection
on the figure axis $x'_3$, but allows for a range of possible angles
$\theta_0$ as long as the top spins fast enough.  If the angular momentum
$p_\psi$ is much larger than this minimum, the square root in
(\ref{eq:precessionbeta}) can be approximated, 
\begin{equation}
\sqrt{1-\frac{4MglI_1\cos\theta_0}{p_\psi^2}}\approx1-\frac{2MglI_1\cos\theta_0}{p_\psi^2}\,,
\end{equation}
leading to two approximate values for $\beta$ of
\begin{equation}\label{eq:approximatebeta}
\beta_+\approx\frac{p_\psi\sin^2\theta_0}{\cos\theta_0}\,,\quad{\rm and}\quad
\beta_-\approx\frac{MglI_1\sin^2\theta_0}{p_\psi}\,.
\end{equation}
For the change of the orientation of the figure axis in real
space, i.e., the precession speed, we use (\ref{eq:otherprecession1}), and
find two values corresponding to the two values for $\beta$ in
(\ref{eq:approximatebeta}): \begin{eqnarray}
\dot{\phi}_+&=&\frac{\beta_+}{I_1\sin^2\theta_0}\approx\frac{p_\psi}{I_1\cos\theta_0}=\frac{I_3\omega'_3}{I_1\cos\theta_0}\,,\nonumber\\[1em]
\dot{\phi}_-&=&\frac{\beta_-}{I_1\sin^2\theta_0}\approx\frac{Mgl}{p_\psi}=\frac{Mgl}{I_3\omega'_3}\,.
\end{eqnarray}
The lower precession rate $\dot{\phi}_-$ is apparently what is typically
observed for this problem, leading to a faster and faster precession rate as a
top slows down.

\clearpage
\subsection{Stability of rotations of rigid bodies}
Even the very simple case of a force-free rotation of a complex object
around its principal rotation axes is not as simple as it appears; we consider
a rigid body with a coordinate system aligned to its principal rotation axes,
assuming $I_3>I_2>I_1$:

\begin{figure*}[h]
\centerline{\includegraphics{rigidbody.eps}}
\end{figure*}
If the body rotates approximately but not exactly around the $x'_1$ axis,
the angular velocity vector can be written as
\begin{equation}
\vec{\omega}=\omega_1\vec{e'}_1+\lambda\vec{e'}_2+\mu\vec{e'}_3\,,\quad{\rm
  with}\quad\lambda, \mu\ll\omega_1\,.
\end{equation}
Euler's equations of motion for the force-free rotating body
(\ref{eq:eulerequationsrigid}) allow  determining the time-dependence of the
coefficients $\omega_1$, $\lambda$ and $\mu$:
\begin{eqnarray}
(I_2-I_3)\lambda\mu-I_1\dot{\omega}_1&=&0\,,\label{eq:rotstab1}\\
(I_3-I_1)\mu\omega_1-I_2\dot{\lambda}&=&0\,,\\
(I_1-I_2)\lambda\omega_1-I_3\dot{\mu}&=&0\,.
\end{eqnarray}
Since we start out with small perturbations of the rotation vector from the
principal axis, we can approximate the product $\lambda\mu$ in
(\ref{eq:rotstab1}) initially by 0, so the angular velocity $\omega_1$ remains
constant over time:
\begin{equation}
-I_1\dot{\omega}_1=0\quad\rightarrow\quad\omega_1=const.
\end{equation}
The other two equations lead to a coupled system of differential equations,
\begin{equation}\label{eq:omegadeviation0}
\dot{\lambda}=\frac{I_3-I_1}{I_2}\omega_1\mu\,,\quad
\dot{\mu}=\frac{I_1-I_2}{I_3}\omega_1\lambda\,.
\end{equation}
We can not use the same trick as for the force-free symmetric top in section
\ref{ssec:forcefreetop}, because the coupling constant is not the
same. Instead, we can chain the equations, effectively eliminating the
variable $\mu$:
\begin{equation}
\ddot{\lambda}=\frac{I_3-I_1}{I_2}\omega_1\dot{\mu}=\frac{I_3-I_1}{I_2}\,\frac{I_1-I_2}{I_3}\omega_1^2\lambda
\end{equation}
or
\begin{equation}\label{eq:omegadeviation}
\ddot{\lambda}\,+\,\Omega_{1}^2 \lambda=0\quad{\rm with}\quad \Omega_{1}:=\omega_1\sqrt{\frac{(I_1-I_3)}{I_2}\,\frac{(I_1-I_2)}{I_3}}\,.
\end{equation}
This second order linear differential equation for $\lambda(t)$ has the familiar solution
\begin{equation}\label{eq:omegadeviation2}
\lambda(t)=Ae^{i\Omega_1 t}+Be^{-i\Omega_1 t}\,.
\end{equation}
For $I_1<I_3$ and $I_1<I_2$, as postulated initially, the angular frequency
$\Omega_1$ is real, resulting in an oscillatory evolution of the rotation
components along axes $x'_2$ and $x'_3$ {\em with a fixed amplitude}. This
means that small deviations of $\vec{\omega}$ from the principal axis $x'_1$
remain small.

The situation changes if the rotation takes place around axis $x'_2$; by
cyclic permutation from the expression in (\ref{eq:omegadeviation}) one obtains
\begin{equation}
\Omega_{2}=\omega_2\sqrt{\frac{(I_2-I_1)}{I_3}\,\frac{(I_2-I_3)}{I_1}}\,.
\end{equation}
This time, the expression in the square root becomes negative, resulting in an
imaginary $\Omega_2$ and therefore an exponentially growing contribution in
the equivalent solution to 
(\ref{eq:omegadeviation2}). Thus, a small deviation of the angular velocity
vector from axis $x_2$ does not lead to a stable rotation.
For a rotation around axis $x'_3$ with the largest moment of inertia, one
obtains 
\begin{equation}
\Omega_{3}=\omega_3\sqrt{\frac{(I_3-I_2)}{I_1}\,\frac{(I_3-I_1)}{I_2}}\,,
\end{equation}
which is real-valued again, resulting in a stability of small perturbations of
the rotation axis alignment with $x'_3$, similar to the rotations around
$x'_1$. In summary, rotations around the axis with the largest and
with the smallest moment of inertia are stable, a rotation around the third
axis is {\em unstable} for a rigid body with three different
principal moments of inertia.

\subsubsection{Symmetric top}
For a symmetric top, with $I_1=I_2$ a rotation around axis $x'_1$  can be
treated similarly, but the equation of motion for $\mu$ in
(\ref{eq:omegadeviation0}) becomes simpler:
\begin{equation}
\dot{\mu}=0\quad\rightarrow\quad\mu=const.
\end{equation}
consequently, the equation for $\lambda$ leads to a non-oscillating solution
for the perturbation component,
\begin{equation}
\dot{\lambda}=\frac{I_3-I_1}{I_2}\omega_1\mu=const.\quad\rightarrow
\lambda(t)=At+B\,,
\end{equation}
which is not limited over time. Thus, the rotation of a symmetric top around an
axis other than the figure axis is {\em unstable}, and only the rotation
around the figure axis $x'_3$ is stable.
\end{document}

%%% Local Variables: 
%%% mode: latex
%%% TeX-master: t
%%% End: 
 
