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\begin{titlepage} \setcounter{page}{0} \vspace{2.5cm}
\begin{center}
\Large\bf{Special Relativity and Maxwell Equations}\\
\vspace{2.0cm} \large{by}\\ \vspace{2.0cm}
\large{\bf Bernd A. Berg }\\ \vspace{2.0cm}
{\bf Department of Physics}\\ \vspace{0.4cm}
{\bf Florida State University}\\ \vspace{0.4cm}
{\bf Tallahassee, FL~32306, USA. }\\ \vspace{0.4cm}
{(Version March 11, 2011)} \vspace{2.0cm} \vfill

{\bf Copyright \copyright\ by the author.}\\

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\chapter{}\label{Relativity}

% \chapter{Special Relativity and Maxwell Equations
% (by B.A. Berg)} \label{Relativity} ! For PHY4241 (W/S 04, 09, 10, 11).

An introduction to the theory of special relativity is given, which 
provides the space-time frame for classical electrodynamics.
Historically~\cite{Ei1905} special relativity emerged
out of electromagnetism. Nowadays, is deserves to be emphasized
that special relativity restricts severely the possibilities for
electromagnetic equations. 

\section{Special Relativity}

Let us deal with space and time in vacuum. The conventional {\it time 
unit} is the
\begin{equation} \label{second}
  second~[s].
\end{equation}
Here and in the following abbreviations for units are placed in 
brackets~$[~]$. For most of the 20th century the second was defined 
in terms of the rotation of the earth as ${1\over 60}\times {1\over 60}
\times {1\over 24}$ of the mean solar day. Nowadays most accurate time 
measurements rely on atomic clocks. They work by tuning an electric 
frequency into resonance with an atomic transition. The second has 
been defined, so that the frequency of the light between the two 
hyperfine levels of the ground state of the cesium $^{132}Cs$ atom 
is exactly 9,192,631,770 cycles per second. \medskip

Special relativity is founded on two basic postulates:

\begin{description}
\item{{\bf 1.}} Galilee invariance: The laws of nature are independent of 
any uniform, translational motion of the reference frame.
\end{description}

This postulate gives rise to a triple infinite set of reference frames 
moving with constant velocities relative to one another. They are called 
{\it inertial frames}. For a freely moving body, i.e., a body which is not 
acted upon by an external force, inertial systems exist. The differential 
equations which describe physical laws take the same form in all inertial 
frames (form invariance). {\it Galilee invariance} was known long before 
Einstein. 

\begin{description}
\item{{\bf 2.}} The speed $c$ of light in empty space is independent 
of the motion of its source.
\end{description}

The second Postulate was introduced by Einstein 1905~\cite{Ei1905}. It
implies that $c$ takes the same constant value in all inertial frames.
Transformations between inertial frames are implied, which have far 
reaching physical consequences.

The {\it distance unit}
\begin{equation} \label{meter}
 1~meter~ [m] = 100~centimeters~[cm] 
\end{equation}
was originally defined by two scratches on a bar made of 
platinum--iridium alloy kept at the International Bureau of Weights 
and Measures in S\`evres, France. As measurements of the speed of 
light became increasingly accurate, Postulate~2 has been exploited
to define the distance unit. The meter is now defined \cite{Petley}
as the distance traveled by light in empty space during the time of 
1/299,792,458~$[s]$. This makes the speed of light exactly 
\begin{equation} c=299,792,458~[m/s] .\label{light}\end{equation}

\subsection{Natural Units}

The units for $second$~(\ref{second}) and $meter$~(\ref{meter}) are
not independent, as the speed of light is a universal constant. This 
allows to define {\it natural units}, which are frequently used in 
nuclear, particle and astro physics. They define 
\begin{equation} \label{natural_units}
  c=1
\end{equation}
as a dimensionless constant, so that
$$ 1~[s] = 299,792,458~[m]  $$
holds. The advantage of natural units is that factors of $c$ disappear
in calculations. The disadvantage is that for converting back to 
conventional units the appropriate factors have to be recovered by 
dimensional analysis. For instance, if time is given in seconds $x=t$
in natural units converts to $x=ct$ with $x$ in meters and $c$ given 
by~(\ref{light}).

\subsection{Definition of distances and synchronization of clocks}

Let us use the concepts of Galilee invariance and of a constant 
speed of light to reduce measurements of spatial distances to time 
measurements. We consider an inertial frame $K$ with coordinates 
$(t,\vec{x})$ and place observers at rest at different places $\vec{x}$ 
in $K$. The observers are equipped with clocks of identical making, 
to define the time $t$ at $\vec{x}$. The origin $(t,\vec{0})$ of $K$ 
is defined by placing an observer $O_0$ with a clock at $\vec{x}=0$.
We like to place another observer $O_1$ at $\vec{x}_1$ to define 
$(t,\vec{x}_1)$. How can $O_1$ know to be at $\vec{x}_1$? By using 
a mirror he can reflect light flashed by observer $O_0$ at him. 
Observer $O_0$ measures the polar and azimuthal angles $(\theta, 
\phi)$ at which he emits the light and 
$$ |\vec{x}_1| = c\, \triangle t / 2\, , $$ 
where $\triangle t$ is the time light needs to travel to $O_1$ and 
back. This determines $\vec{x}_1$ and $O_0$ signals this information 
to $O_1$. By repeating the measurement, he can make sure that $O_1$ is 
not moving with respect to $K$. For an idealized, force free environment 
the observers will then never start moving with respect to one another. 
$O_1$ synchronizes her clock by setting it to
$$ t_1^r=t_1^e+|\vec{x}_1|/c $$
where $O_0$ {\it emits} (superscript $e$) the signal at $t_1^e$ and 
$O_1$ {\it receives} (superscript $r$) it at $t^r_1$. When $O_0$ 
flashes later his instant time $t^e_2$ over to $O_1$, the clock of 
$O_1$ will show time $t^r_2=t^e_2+|\vec{x}_1|/c$ when receiving the 
signal. In the same way the time $t$ can be defined at any desired 
point $\vec{x}$ in $K$. 

Now we consider an inertial frame $K'$ with coordinates $(t',\vec{x}^{
\,'})$, moving with {\it constant velocity} $\vec{v}$ with respect to 
$K$. The origin of $K'$ is defined through an observer $O'_0$. How does
one know that $O'_0$ moves with constant velocity $\vec{v}$ with respect 
to $O_0$? At times $t^e_1$ and $t^e_2$ observer $O_0$ may flash light 
signals at $O'_0$, which are reflected and arrive back after time 
intervals $\triangle t_1$ and $\triangle t_2$ on the clock of $O_0$. 
From principle~2 it follows that the reflected light needs the same 
time to travel from $O'_0$ to $O_0$, as it needed to travel from $O_0$ 
to $O'_0$. Hence, $O_0$ concludes that $O_0'$ received the signals at
\begin{equation}
  t^r_i=t^e_i+\triangle t_i/2,\ (i=1,2) 
\end{equation}
in the $O_0$ time. This simple equation becomes more complicated for 
non-re\-la\-ti\-vistic physics, because the speed on the return path 
would then be distinct from that on the arrival path (consider elastic 
scattering of a very light particle on a heavy surface). The constant 
velocity of light implies that relativistic distance measurements are 
simpler than such non-relativistic measurements. For observer $O_0$ the 
vector positions $\vec{x}_1$ and $\vec{x}_2$ of $O_0'$ at times $t^r_1$ 
and $t^r_2$, respectively, are completely defined by the angles 
$(\theta_i,\phi_i)$ at which the light comes back and the magnitude
\begin{equation}
  |\vec{x}_i| = \triangle t_i\, c/2,\ (i=1,2)\, .
\end{equation}
For the assumed force free environment observer $O_0$ can conclude 
that $O'_0$ moves with respect to him with uniform velocity 
\begin{equation}
\vec{v} = (\vec{x}_2-\vec{x}_1) / (t^r_2-t^r_1)\, .
\end{equation}
Actually, one measurement is sufficient to obtain the velocity when
one employs the relativistic Doppler effect as discussed later in 
section~\ref{sec_Doppler}. $O_0$ may repeat the procedure to check 
that $O_o'$ moves indeed with uniform velocity.

The equation of motion for the origin of $K'$ as observed by $O_0$ is
\begin{equation} \label{O'_0}
  \vec{x}\, (\vec{x}^{\, '}=0) = \vec{x}_0 + \vec{v}\, t , 
\end{equation}
with $\vec{x}_0=\vec{x}_1-\vec{v}\,t^r_1$, expressing the fact 
that for $t=t^r_1$ observer $O'_0$ is at $\vec{x}_1$. Shifting his 
space convention by a constant vector, observer $O_0$ can achieve 
$\vec{x}_0=0$, so that equation (\ref{O'_0}) becomes
$$ \vec{x}\, (\vec{x}^{\, '}=0) = \vec{v}\, t . $$
Similarly, observer $O'_0$ finds out that $O_0$ moves with velocity 
$\vec{v}{\, '}=-\vec{v}$. According to principle~1, observers in $K'$ 
can now go ahead to define $t'$ for any point $\vec{x}{\, '}$ in $K'$. 
Observer $O_0'$ can choose his space convention so that
$$ \vec{x}^{\, '}\, (\vec{x}=0) = - \vec{v}\, t' $$
holds. 

\subsection{Lorentz invariance and Minkowski space}

Having defined time and space operationally, let us focus on a more 
abstract discussion. We consider two inertial frames with uniform 
relative motion $\vec{v}$: $K$ with coordinates $(t,\vec x)$ and $K'$ 
with coordinates $(t',\vec{x}{\, '})$. We demand that at time $t=t'=0$ 
their two origins coincide. Now, imagine a spherical shell of radiation 
originating at time $t=0$ from $\vec{x}=\vec{x}{\, '}=0$. The 
propagation of the wavefront is described by
\begin{equation}
  c^2 t^2 - x^2 - y^2 - z ^2 = 0 ~~~{\rm in}\ K ,  \label{inK}
\end{equation}
and by
\begin{equation}
  c^2 t'^{\,2}-x'^{\,2}-y'^{\,2}-z'^{\,2}= 0~~~{\rm in}\ K' .\label{inKP} 
\end{equation}
We define 4-vectors $(\alpha = 0,1,2,3)$ by
\begin{equation} \label{4x}
  (x^{\alpha})=\left( \matrix{ct\cr  \vec{x}\cr} \right) 
  ~~{\rm and}~~(x_{\alpha})=\left( ct,\ -\vec{x} \right)\, . 
\end{equation}
Due to a more general notation, which is explained in 
section~\ref{sec_tensor}, the components $x^{\alpha}$ are called 
{\it contravariant} and the components $x_{\alpha}$ {\it covariant}. 
In matrix notation the contravariant 4-vector $(x^{\alpha})$ is 
represented by a column and the covariant 4-vector $(x_{\alpha})$ 
as a row. The contravariant vectors are those encountered in
non-relativistic mechanics, where unfortunately the indices are 
usually down instead of up.

The {\it Einstein summation convention} is defined by
\begin{equation} \label{Einsum}
  x_{\alpha} x^{\alpha} = \sum_{\alpha=0}^3 x_{\alpha} x^{\alpha}
  = (x^0)^2 -\vec{x}^{\,2}\, ,
\end{equation}
and will be employed from here on. Equations (\ref{inK}) and 
(\ref{inKP}) read then 
\begin{equation} \label{InvLight}
  x_{\alpha}x^{\alpha} = x'_{\, \alpha} x'^{\, \alpha} = 0\ . 
\end{equation}
{\it (Homogeneous) {Lorentz transformations}} are defined as the group 
of transformations which leave the distance $s^2=x_{\alpha}x^{\alpha}$
invariant:
\begin{equation} \label{HLo}
  x_{\alpha}x^{\alpha} = x'_{\alpha}x'^{\,\alpha} = s^2\ .
\end{equation}
This equation implies (\ref{InvLight}), but the reverse is not true.
An additional transformation, which leaves (\ref{InvLight}), but not
(\ref{HLo}) invariant, is the scale transformation $x'^{\,\alpha} =
\lambda\,x^{\alpha}$.

If the initial condition $t'=0$ and $\vec{x}^{\,'}(\vec{x}=0)
=\vec{x}(\vec{x}^{\,'})=0$ for $t=0$ is replaced by an arbitrary one, 
the equation $(x_{\alpha}-y_{\alpha})(x^{\alpha}-y^{\alpha}) =
 (x'_{\,\alpha}-y'_{\,\alpha})(x'^{\alpha}-y'^{\alpha}) $ still holds.
\index{}{\it Inhomogeneous Lorentz} or 
\index{}{\it Poincar\'e} transformations are
defined as the group of transformations which leave
\begin{equation} \label{ILo}
s^2=(x_{\,\alpha}-y_{\,\alpha})(x^{\alpha}-y^{\alpha})
  ~~~{\rm invariant.} 
\end{equation}
In contrast to the Lorentz transformations the Poincar\'e 
transformations include invariance under 
\index{}{\it translations} 
\begin{equation} \label{translations}
  x^{\alpha} \to x^{\alpha} + a^{\alpha} ~~{\rm and}~~
  y^{\alpha} \to y^{\alpha} + a^{\alpha} 
\end{equation}
where $a^{\alpha}$ is a constant vector. Independently of Einstein, 
Poincar\'e had developed similar ideas, but pursued a more cautious 
approach.

\begin{figure}[tb] \vspace{5pc}
\centerline{\hbox{ \psfig{figure=fig/Minkowski.eps,width=10cm} }}
\caption{Minkowski space: Seen from the spacetime point A at the origin,
the spacetime points in the forward light cone are in the future, those 
in the backward light cone are in the past and the spacelike points are 
``elsewhere'', because their time-ordering depends on the inertial 
frame chosen. Paths of two clocks which separate at the origin (the 
straight line one stays at rest) and merge again at a future space-time 
point B are also indicated. For the paths shown the clock moved along 
the curved (in the figure longer!) path will, at B, show an elapsed 
time of about 70\% of the elapsed time shown by the other clock, which 
stays at rest.  \label{fig_Minkowski} }
\end{figure}

A fruitful concept is that of a 4-dimensional space-time, 
called {\it Minkowski space}. Equation (\ref{ILo}) gives the 
invariant metric of this space. Compared to the 
norm of 4-dimensional Euclidean space, the crucial difference 
is the relative minus sign between time and space components.
The \index{}{\it light cone} of a 4-vector $x^{\alpha}_0$ is 
defined as the set of vectors $x^{\alpha}$ which satisfy
$$ (x-x_0)^2 =
(x_{\alpha}-x_{0\alpha})\, (x^{\alpha}-x^{\alpha}_0) = 0 .$$
The light cone separates events which are \index{}{\it timelike} 
and \index{}{\it spacelike} with respect to $x^{\alpha}_0$, 
namely
$$ (x-x_0)^2 > 0 ~~{\rm for\ timelike} $$
and
$$ (x-x_0)^2 < 0 ~~{\rm for\ spacelike.} $$
We shall see soon, note equation~(\ref{2DLT1}), that the 
{\it time-ordering} of spacelike points is distinct in different 
inertial frames, whereas it is the same for timelike points. For the 
choice $x_0^{\alpha}=0$ this Minkowski space situation is depicted 
in figure~\ref{fig_Minkowski}. On the abscissa we have the projection 
of the three dimensional Euclidean space on $r=|\vec{x}|$. The regions
\index{} {\it future} and \index{} {\it past} of this figure are 
the timelike points of $x_0=0$, whereas {\it elsewhere} are the
spacelike points.

To understand special relativity in some depth, we have to explore 
Lorentz and Poincar\'e transformations. Before we come to this, 
we consider the two-dimensional case and introduce some relevant 
calculus in the next two sections.

\subsection{Two-Dimensional Relativistic Kinematics}

We chose now $\vec{v}$ in $x$-direction and restrict the discussion
to the $x$-axis: 
\begin{equation}
 c^2 t^2 - (x^1)^2 = c^2 t'^2 -(x^{'\, 1})^2 \label{2Dinvar}
\end{equation}
As before, $x^0=ct$, $x^{'\, 0}=ct'$ and $\beta=v/c$. We are looking 
for a linear transformation
\begin{equation}
  \pmatrix{x^{'\,0}\cr x^{'\,1}\cr} = \pmatrix{a&b\cr d&e\cr}
  \pmatrix{x^0\cr x^1\cr} \label{LTansatz}
\end{equation}
which fulfills (\ref{2Dinvar}) for all $x^0$, $x^1$. Choosing 
$\pmatrix{x^0\cr x^1\cr} = \pmatrix{1\cr 0\cr}$ gives
\begin{equation} \label{a00} 
  a^2-d^2=1\ \Rightarrow\ a=\cosh(\zeta ),\ d=\pm\sinh(\zeta )
\end{equation}
and choosing $\pmatrix{x^0\cr x^1\cr} = \pmatrix{0\cr 1\cr}$ gives
\begin{equation} \label{a11}
  b^2-e^2=-1\ \Rightarrow\ e=\cosh(\eta),\ b=\pm\sinh(\eta)\ . 
\end{equation}
Using now $\pmatrix{x^0\cr x^1\cr}=\pmatrix{1\cr 1\cr}$
yields 
$$ [\cosh (\zeta ) + \sinh (\eta ) ]^2 -
   [\sinh (\zeta ) + \cosh (\eta ) ]^2 = 0\
\Rightarrow\ \zeta = \eta . $$
In equation (\ref{a00}) we choose the usual convention $d=-\sinh(\zeta
)$ and end up with \begin{equation} \label{hyperbolic}
  \pmatrix{a&b\cr d& e\cr} = \pmatrix{\cosh(\zeta )&-\sinh(\zeta)
  \cr -\sinh(\zeta)& \cosh(\zeta)\cr} ,
\end{equation}
where $\zeta$ is called {\it rapidity} or {\it boost} variable and 
has the interpretation of an angle in a hyperbolic geometry. For our 
present purposes no knowledge of hyperbolic geometries is required. 
In components (\ref{LTansatz}) reads then
\begin{eqnarray} \label{2DLT1}
   x'^{\,0} &=& +\cosh(\zeta)\,x^0 -\sinh(\zeta)\,x^1\,,\\
   x'^{\,1} &=& -\sinh(\zeta)\,x^0 +\cosh(\zeta)\,x^1\,. \label{2DLT2}
\end{eqnarray} 
An interesting feature of equation (\ref{2DLT1}) is that for spacelike
points, say $x^1>x^0>0$, a value $\zeta_0$ for the rapidity exists, so 
that 
$$ 0 = +\cosh(\zeta)\,x^0 -\sinh(\zeta)\,x^1 $$
and, therefore,
$$ {\rm sign} (x'^{\, 0}) = - {\rm sign} (x^0) $$ 
for $\zeta > \zeta_0$, so that the time-ordering becomes reversed, 
whereas for timelike points such a reversal of the time-ordering 
is impossible as then $|x^0|>|x^1|$. In figure~\ref{fig_Minkowski} 
this is emphasized by calling the spacelike (with respect to $x_0=0$) 
region {\it elsewhere} in contrast to {\it future} and {\it past}.

The physical interpretation is straightforward. Seen from $K$, the 
origin $x'^{\,1}=0$ of $K'$ moves with constant velocity $v$. In 
$K$ this corresponds to the equation 
$$ 0 = -\sinh(\zeta)\,x^0 +\cosh(\zeta)\,x^1 $$
and the rapidity is related to the velocity between the frames by
\begin{equation} \label{rapidity}
  \beta = {v\over c} = {x^1\over x^0} = 
  {\sinh(\zeta)\over \cosh(\zeta)} = \tanh(\zeta)\ . 
\end{equation}
Another often used notation is
\begin{equation} \label{gamma}
  \gamma = \cosh (\zeta ) = {1\over\sqrt{1-\beta^2}}~~{\rm and}
  ~~\gamma\,\beta = \sinh (\zeta )\ . 
\end{equation}
Hence, the transformation (\ref{LTansatz}) follows in 
the often stated form
\begin{eqnarray} \label{LT1a}
  x^{'\,0} &=& \gamma\,(x^0-\beta\,x^1)\, ,\\ \label{LT2a}
  x^{'\,1} &=& \gamma\,(x^1-\beta\,x^0)\, .
\end{eqnarray} 
These equation are called {\it Lorentz transformations}. Lorentz 
discovered them first in his studies of electrodynamics, but it 
remained due to Einstein \cite{Ei1905} to understand their physical 
meaning. We may perform two subsequent Lorentz transformations with 
rapidity $\zeta_1$ and $\zeta_2$. They combine as follows:
$$ \pmatrix{+\cosh (\zeta_2 ) & -\sinh (\zeta_2 )\cr 
           -\sinh (\zeta_2 ) & +\cosh (\zeta_2 )\cr}\
  \pmatrix{+\cosh (\zeta_1 ) & -\sinh (\zeta_1 )\cr 
           -\sinh (\zeta_1 ) & +\cosh (\zeta_1 )\cr}  $$
\begin{equation}
=\ \pmatrix{+\cosh (\zeta_2+\zeta_1 ) & -\sinh (\zeta_2+\zeta_1 )\cr 
            -\sinh (\zeta_2+\zeta_1 ) & +\cosh (\zeta_2+\zeta_1 )\cr}\ .
\end{equation}
The rapidities add up as
\begin{equation} \label{rapidity_add} 
  \zeta = \zeta_1 + \zeta_2
\end{equation}
in the same way as velocities under Galilei transformations or angles 
for rotations about the same axis do. Note that the inverse to the 
transformation with rapidity $\zeta_1$ is obtained for $\zeta_2 =
-\zeta_1$. The relativistic addition of velocities follows from 
(\ref{rapidity_add}). Let $\beta_1=\tanh(\zeta_1)$ and
$\beta_2=\tanh(\zeta_2)$, then
\begin{equation} \label{velocity_add} 
  \beta = \tanh(\zeta_1+\zeta_2) = 
  \frac{\beta_1+\beta_2}{1+\beta_1\beta_2}
\end{equation}
holds. Another immediate consequence of the Lorentz transformations is 
the {\it time dilatation}: A moving clock ticks slower. In $K$ the 
position of the origin of $K'$ is given by 
$$ x^1=v\,x^0/c = \tanh(\zeta)\,x^0 $$
and the Lorentz transformation (\ref{2DLT1}) gives
\begin{equation} \label{Tdil}
  x^{'\, 0} = \cosh(\zeta)\,x^0 - \sinh(\zeta)\,\tanh(\zeta)\,x^0
            = \frac{\cosh^2(\zeta)-\sinh^2(\zeta)}{\cosh(\zeta)}\,x^0
            = \frac{x^0}{\cosh(\zeta)} < x^0\ .
\end{equation} 
This works also the other way round. In $K'$ the position of the 
origin of $K$ is given by 
$$ x^{'\, 1} = -\tanh(\zeta)\, x^{'\, 0} $$
and with this relation between $x^{'\, 1}$ and $x^{'\, 0}$ the
inverse Lorentz transformation gives
$$ x^0=x^{'\, 0} / \cosh(\zeta)\ .$$
There is no paradox, because equal times at separate points in one 
frame are not equal in another (remember that the definition of time 
in one frame relies already on the constant speed of light). In particle 
physics the effect is day by day observed for the lifetimes of unstable 
particles. To test time dilatation for macrosciopic clocks, we have to 
send a clock on a roundtrip. For this an infinitesimal form of equation 
(\ref{Tdil}) is needed. 

Allowing that $x^0=x^1=0$ does not have to 
coincide with $x^{'\,0} =x^{'\, 1}=0$, we consider 
Poincar\'e transformations. The light radiation may originate in $K$ 
at $(x^0_0, x^1_0)$ and in $K'$ at $(x^{'\, 0}_0,x^{'\, 1}_0)$. This 
generalizes equation (\ref{2Dinvar}) to 
$$ (x^{'\, 0}-x^{'\, 0}_0)^2 - (x^{'\, 1}-x^{'\, 1}_0)^2 = 
   (x^0-x^0_0)^2 - (x^1-x^1_0)^2 , $$
and the Lorentz transformations become
\begin{equation} 
  (x^{'\, 0} -  x^{'\, 0}_0) = \gamma\, 
  [(x^0 -x^0_0) - \beta\, (x^1-x^1_0)]\, , \label{Po1}
\end{equation} 
\begin{equation} 
 (x^{'\, 1} - x^{'\, 1}_0) = 
\gamma\, [(x^1-x^1_0) - \beta\, (x^0-x^0_0)]\, . \label{Po2}
\end{equation} 
Using the rapidity variable and matrix notation:
\begin{equation} \label{Po3}
\pmatrix{ x^{'\, 0}-x^{'\, 0}_0\cr  x^{'\, 1}-x^{'\, 1}_0 \cr} 
=  \pmatrix{ \cosh (\zeta )& -\sinh (\zeta )\cr 
  -\sinh (\zeta )& \cosh (\zeta ) \cr}\
   \pmatrix{ x^0-x^0_0\cr  x^1-x^1_0\cr}\ .
\end{equation} 
In addition we have invariance under translations (\ref{translations}).

Let us explore Minkowski space in more details. It allows one to depict 
{\it world lines} of particles. A useful concept for a particle (or 
observer) traveling along its world line is its {\it proper time}
or {\it eigenzeit}. Assume the particle moves with velocity $v(t)$, 
then $dx^1=\beta\,dx^0$ holds, and the infinitesimal invariant along 
its 2D world line is
\begin{equation} \label{2Dworld_line}
  (ds)^2 = (dx^0)^2-(dx^1)^2 = (c\,dt)^2\,(1-\beta^2)\ . 
\end{equation} 
Each instantaneous rest frame of the particle is an inertial frame. 
The increment of time $d\tau$ in such an instantaneous rest frame 
is a Lorentz invariant quantity which takes the form
\begin{equation} \label{dtau}
 d\tau = dt\, \sqrt{1-\beta^2} = dt\, \gamma^{-1}
       = dt/\cosh \zeta \ ,
\end{equation} 
where $\tau$ is called proper time. Clocks click by their proper time.
As $\gamma (\tau) \ge 1$ time dilatation follows
\begin{equation} \label{clocktrip}
  \int_{t_1}^{t_2}dt =  t_2-t_1 = \int_{\tau_1}^{\tau_2} \gamma(\tau)
  \,d\tau = \int_{\tau_1}^{\tau_2} \cosh\zeta(\tau)\,d\tau \ge 
  \tau_2 - \tau_1\, . 
\end{equation} 
A moving clock runs more slowly than a stationary clock. 
Equation~(\ref{clocktrip}) applies to general paths of a clock, 
including those with acceleration. Relevant is that the 
entire derivation was done with respect to the inertial system 
in which the time coordinates $t_2$ and $t_1$ are defined.
Two experimental examples for time dilatation are: (i)~Time of flight
of unstable particles in high energy scattering experiments, where 
these particles move at velocities close to the speed of light.
(ii)~Explicit verification through travel with atomic clocks on 
air planes \cite{Hef1,Vessot}. 

Next, let us discuss the relation between velocity and {\it 
acceleration}.  Assume an accelaration $a$ in the instanteneous 
rest frame. To have convenient units we define $\alpha =a/c$ and
\begin{equation} \label{2Dacceleration}
  d\beta = d\zeta = \alpha\,d\tau
\end{equation} 
holds in the instantenous rest frame. The change in another frame
follow from the addition theorem of velocities (\ref{velocity_add})
\begin{equation} \label{dbeta}
  d\beta = \frac{\alpha d\tau+\beta}{1+\alpha d\tau\,\beta}-\beta
         = \alpha\,(1-\beta^2)\,d\tau\,.
\end{equation} 
As rapidities simply add up (\ref{rapidity_add}) their change is just
\begin{equation} \label{dzeta}
  d\zeta = (\alpha\,d\tau + \zeta) - \zeta = \alpha\,d\tau\ .
\end{equation} 
Using the proper time, the change of the rapidity is analogue to the 
change of the velocity in non-relativistic mechanics, 
\begin{equation} \label{zeta_tau}
  \zeta - \zeta_0 = \int_{\tau_0}^{\tau} \alpha(\tau)\,d\tau\ .
\end{equation} 
In particular, if $\alpha$ is constant we can integrate and find
\begin{equation} \label{zeta_const}
  \zeta(\tau) = \alpha \tau + \zeta_0~~~{\rm for}~~\alpha~~
  {\rm constant}\ .
\end{equation} 

\subsection{Vector and tensor notation\label{sec_tensor}}

One defines a general transformation $x\to x'$ through
\begin{equation} \label{trans}
x'^{\, \alpha} = x'^{\, \alpha} (x) =
x'^{\, \alpha}\,\left( x^0, x^1, x^2, x^3 \right),\ \alpha = 0,1,2,3. 
\end{equation}
This means, $x'^{\,\alpha}$ is a function of four variables and,
when it is needed, this function is assumed to be sufficiently 
often differentiable with respect to each of its arguments. In the 
following we consider the transformation properties of various
quantities (scalars, vectors and tensors) under $x\to x'$.

A {\it scalar} is a single quantity whose value is not changed under 
the transformation (\ref{trans}). The proper time is an example.

A {\it 4-vector} $A^{\alpha}, (\alpha=0,1,2,3)$ is said 
{\it contravariant} if its components transform according to 
\begin{equation}
  A'^{\alpha} = {\partial x'^{\alpha} \over \partial x^{\beta}} 
  A^{\beta} . \label{con}
\end{equation}
An example is $A^{\alpha} = dx^{\alpha}$, where (\ref{con})
reduces to the well--known rule for the differential of a
function of several variable $(f^{\alpha}(x)=x'^{\alpha}(x))$:
$$ dx'^{\alpha} = {\partial x'^{\, \alpha} \over \partial x^{\beta}} 
   dx^{\beta} . $$
Remark: In this general framework the vector $x^{\alpha}$ itself is 
not always contravariant (for a discussion see books on {\it General 
Relativity} like \cite{Weinberg}). When a linear transformation
$$ x'^{\,\alpha} = a^{\alpha}_{\ \beta}\, x^{\,\beta} $$
holds with space-time independent coefficients $a^{\alpha}_{\ \beta}$, 
then $x^{\,\alpha}$ is contravariant and one finds
$$ {\partial x'^{\,\alpha}\over\partial x^{\,\beta}} =
   a^{\alpha}_{\ \beta} \ .$$
In special relativty we are only interested in linear transformations. 
Space-time dependent transformations lead into general relativity.

A 4-vector is said {\it covariant} when it transforms like
\begin{equation}
  B'_{\alpha} = {\partial x^{\beta} \over \partial x'^{\alpha}} 
  B_{\beta} . \label{cov}
\end{equation}
An example is 
\begin{equation}
  B_{\alpha} = \partial_{\alpha} = 
  {\partial\ \over \partial x^{\alpha}} , \label{partial}
\end{equation}
because of
$$ {\partial\ \over \partial x'^{\alpha}} =
   {\partial x^{\beta} \over \partial x'^{\alpha}}
   {\partial\ \over \partial x^{\beta}} . $$
The {\it inner} or {\it scalar product} of two vectors is defined as
the product of the components of a covariant and a contravariant
vector:
\begin{equation}
 B\cdot A = B_{\alpha} A^{\alpha} . \label{scalarp}
\end{equation}
It follows from (\ref{con}) and (\ref{cov}) that the scalar product is 
an invariant under the transformation (\ref{trans}): 
$$ B' \cdot A' = {\partial x^{\beta} \over \partial x'^{\alpha} }\,
 {\partial x'^{\alpha} \over \partial x^{\gamma} }\, 
 B_{\beta} A^{\gamma} = 
 {\partial x^{\beta} \over \partial x^{\gamma}}\,
 B_{\beta} A^{\gamma} = 
  \delta^{\beta}_{\ \gamma} B_{\beta} A^{\gamma} = B\cdot A .$$
Here the {\it Kronecker delta} is defined by:
\begin{equation} \label{delta}
  \delta^{\alpha}_{\ \beta} = \delta_{\alpha}^{\ \beta} =
  \cases{ 1 ~~{\rm for}~~ \alpha = \beta, \cr
        0 ~~{\rm for}~~ \alpha \ne \beta.} 
\end{equation}
Vectors are {\it rank} one tensors. {\it Tensors} of general rank $k$
are quantities with $k$ indices, like for instance
$$ {T^{\alpha_1\alpha_2 ...}}_{...\alpha_i ... \alpha_k}\ . $$
The convention is that the upper indices transform contravariant
and the lower transform covariant. For instance, a contravariant tensor
of rank two $F^{\alpha\beta}$ consists of 16 quantities that transform
according to
$$ F'^{\alpha\beta} = 
{\partial x'^{\alpha}\over \partial x^{\gamma}} 
{\partial x'^{\beta}\over \partial x^{\delta}} F^{\gamma\delta}\,.$$
A covariant tensor of rank two $G_{\alpha\beta}$ transforms as 
$$ G'_{\alpha\beta} = 
{\partial x^{\gamma}\over \partial x'^{\alpha}} 
{\partial x^{\delta}\over \partial x'^{\beta}} G_{\gamma\delta}\,.$$
The {\it inner product} or {\it contraction} with respect to a pair 
of indices, either on the same tensor or between different tensors,
is defined as in (\ref{scalarp}). One index has to be contravariant 
and the other covariant.

A tensor $S^{...\alpha ...\beta ...}$ is said to be symmetric in 
$\alpha$ and $\beta$ when
$$ S^{...\alpha ...\beta ...} = S^{...\beta ...\alpha ...} . $$
A tensor $A^{...\alpha ...\beta ...}$ is said to be antisymmetric in 
$\alpha$ and $\beta$ when
$$ A^{...\alpha ...\beta ...} = - A^{...\beta ...\alpha ...} . $$
Let $S^{...\alpha ...\beta}$ be a symmetric and $A^{...\alpha ...\beta}
$ be an antisymmetric tensor. It holds
\begin{equation} \label{SA} 
  S^{...\alpha ...\beta ...} A_{...\alpha ...\beta ...} =\ 0 . 
\end{equation}
Proof:
$$ S^{...\alpha ...\beta ...} A_{...\alpha ...\beta ...} = 
 - S^{...\beta ...\alpha ...} A_{...\beta ...\alpha ...} =
 - S^{...\alpha ...\beta ...} A_{...\alpha ...\beta ...} , $$
and consequently zero. The first step exploits symmetry and
antisymmetry, and the second step renames the summation (dummy) 
indices.  Every tensor can be written as a sum of its symmetric 
and antisymmetric parts in two if its indices
\begin{equation} \label{TSA} 
T^{...\alpha ...\beta ...} = T_S^{...\alpha ...\beta ...} 
 + T_A^{...\alpha ...\beta ...} 
\end{equation}
by simply defining
\begin{equation} 
 T_S^{...\alpha ...\beta ...} = {1\over 2} \left(
T^{...\alpha ...\beta ...} + T^{...\beta ...\alpha ...} 
\right) ~~{\rm and}~~
 T_A^{...\alpha ...\beta ...} = {1\over 2} \left(
T^{...\alpha ...\beta ...} - T^{...\beta ...\alpha ...} 
\right)\ . 
\end{equation}

So far the results and definitions are general. We now specialize to 
Poincar\'e transformations. The specific geometry of the space--time 
of special relativity is defined by the invariant distance $s^2$, see 
equation~(\ref{ILo}). In differential form, the infinitesimal interval 
$ds$ defines the proper time $c\,d\tau =ds$,
\begin{equation}
  (ds)^2 = (dx^0)^2 - (dx^1)^2 - (dx^2)^2 - (dx^3)^2\ .
\label{ILo1} \end{equation}
Here we have used superscripts on the coordinates in accordance to our
insight that $dx^{\alpha}$ is a contravariant vector. Introducing a
{\it metric tensor} $g_{\alpha\beta}$ we re--write equation (\ref{ILo1}) as
\begin{equation}
 (ds)^2 = g_{\alpha\beta}\, dx^{\alpha} dx^{\beta} .
\label{ILo2} \end{equation}
Comparing (\ref{ILo1}) and (\ref{ILo2}) we see that for special relativity
$g_{\alpha\beta}$ is diagonal:
\begin{equation} \label{metric_tensor}
g_{00}=1,\, g_{11}=g_{22}=g_{33}=-1 ~~{\rm and}~~
g_{\alpha\beta}=0 ~{\rm for}~ \alpha\ne \beta . \label{gab}
\end{equation}
Comparing (\ref{ILo2}) with the invariant scalar product 
(\ref{scalarp}), we conclude that
$$ x_{\alpha} = g_{\alpha\beta}\, x^{\beta} . $$
The covariant metric tensor lowers the indices, i.e., transforms
a contravariant into a covariant vector. Correspondingly the 
contravariant metric tensor $g^{\alpha\beta}$ is defined to raise
indices:
$$ x^{\alpha} = g^{\alpha\beta}\, x_{\beta} .$$
The last two equations and the symmetry of $g_{\alpha\beta}$ imply
$$ g_{\alpha\gamma}\, g^{\gamma\beta} = \delta_{\alpha}^{\ \beta} $$
for the contraction of the contravariant with the covariant metric 
tensor. This equation yields $g^{\alpha\beta}$, called the normalized 
co--factor of $g_{\alpha\beta}$. For the diagonal matrix (\ref{gab}) 
the result is simply
\begin{equation} \label{metric_tensor_up}
 g^{\alpha\beta} = g_{\alpha\beta} . 
\end{equation}
Raising and lowering indices with $g_{\alpha\beta}$ and 
$g^{\alpha\beta}$, the equations 
$$ A^{\alpha} = \left( \matrix{A^0\cr \vec{A} \cr} \right), 
   ~~~ A_{\alpha} = (A^0, - \vec{A} ) $$
and, compare (\ref{partial}),
\begin{equation} \label{partial2}
 (\partial_{\alpha}) = 
 \left( {\partial \over c\partial t}\,, \nabla \right), ~~~
 (\partial^{\alpha}) = \left( \matrix{ 
  \frac{{\partial}}{{c}\partial{t}}\cr ~\cr 
   - \nabla\cr} \right)\, . 
\end{equation}
% \frac{{{\big\partial}}}{{c}\partial{t}}\cr ~\cr 
are found. It follows that the 4-divergence of a 4-vector
$$ \partial^{\alpha} A_{\alpha} = \partial_{\alpha} A^{\alpha} 
= {\partial A^0 \over \partial x^0} + \nabla \cdot \vec{A}$$
and the d'Alembert (4-dimensional Laplace) operator
$$ \square = \partial_{\alpha} \partial^{\alpha} =
{\left(\partial~~\over \partial x^0\right)^2} - \nabla^2 $$
are invariants. Sometimes the notation $\triangle = \nabla^2$ 
is used for the (3-dimensional) Laplace operator.

\subsection{Lorentz transformations}

Let us now construct the Lorentz group. We seek a group of linear 
transformations
\begin{equation}
 x'^{\alpha} = a^{\alpha}_{\ \beta}\, x^{\beta} ,\ \
(\Rightarrow {\partial x'^{\alpha} \over \partial x^{\beta}}
= a^{\alpha}_{\ \beta} ) \label{Ltrans}
\end{equation}
so that the scalar product stays invariant:
$$ x'_{\alpha}\, x'^{\alpha} = a_{\alpha}^{\ \beta} x_{\beta}\,
a^{\alpha}_{\ \gamma} x^{\gamma} = x_{\alpha}\, x^{\alpha}
 = \delta^{\beta}_{\ \gamma} x_{\beta} x^{\gamma} .$$
As the $x_{\beta} x^{\gamma}$ are independent, this yields
$$ 
 a_{\alpha}^{\ \beta}\, a^{\alpha}_{\ \gamma} = 
 \delta^{\beta}_{\ \gamma}\ \Leftrightarrow \
 a_{\alpha\beta}\, a^{\alpha}_{\ \gamma} = 
 g_{\beta\gamma}\ \Leftrightarrow \
 a^{\delta}_{\ \beta}\, g_{\delta\alpha}\, a^{\alpha}_{\ \gamma} = 
 g_{\beta\gamma}\ . $$
In matrix notation
\begin{equation}
\tilde{A} g A = g , \label{Lg1}
\end{equation}
where $g=(g_{\beta\alpha})$ is given by (\ref{metric_tensor}),
\begin{equation} \label{mat_A}
   A = (a^{\beta}_{\ \alpha}) = \left( \matrix{ 
   a^0_{\ 0} & a^0_{\ 1} & a^0_{\ 2} & a^0_{\ 3}\cr 
   a^1_{\ 0} & a^1_{\ 1} & a^1_{\ 2} & a^1_{\ 3}\cr 
   a^2_{\ 0} & a^2_{\ 1} & a^2_{\ 2} & a^2_{\ 3}\cr 
   a^3_{\ 0} & a^3_{\ 1} & a^3_{\ 2} & a^3_{\ 3}\cr } \right)\ , 
\end{equation}
and $\tilde{A}=(\tilde{a}_{\beta}^{\ \alpha})$ with 
$\tilde{a}_{\beta}^{\ \alpha} = a^{\alpha}_{\ \beta}$ 
is the transpose of the matrix $A=(a^{\beta}_{\ \alpha})$, 
explicitly
\begin{equation} \label{mat_A_transpose}
\tilde{A} = (\tilde{a}_{\beta}^{\ \alpha}) = \left( \matrix{ 
\tilde{a}_0^{\ 0}&\tilde{a}_0^{\ 1}&\tilde{a}_0^{\ 2}&\tilde{a}_0^{\ 3}\cr 
\tilde{a}_1^{\ 0}&\tilde{a}_1^{\ 1}&\tilde{a}_1^{\ 2}&\tilde{a}_1^{\ 3}\cr 
\tilde{a}_2^{\ 0}&\tilde{a}_2^{\ 1}&\tilde{a}_2^{\ 2}&\tilde{a}_2^{\ 3}\cr 
\tilde{a}_3^{\ 0}&\tilde{a}_3^{\ 1}&\tilde{a}_3^{\ 2}&\tilde{a}_3^{\ 3}\cr }
 \right) = \left( \matrix{ 
   a^0_{\ 0}& a^1_{\ 0}& a^2_{\ 0}& a^3_{\ 0}\cr 
   a^0_{\ 1}& a^1_{\ 1}& a^2_{\ 1}& a^3_{\ 1}\cr 
   a^0_{\ 2}& a^1_{\ 2}& a^2_{\ 2}& a^3_{\ 2}\cr 
   a^0_{\ 3}& a^1_{\ 3}& a^2_{\ 3}& a^3_{\ 3}\cr } \right)\ .
\end{equation}
For this definition of the transpose matrix the row indices are 
contravariant and the column indices are covariant, vice verse
to the definition (\ref{4x}) for vectors and, similarly, ordinary
matrices. 
Certain properties of the transformation matrix $A$ can be deduced
from (\ref{Lg1}). Taking the determinant on both sides gives us
$ {\rm det} ( \tilde{A} g A ) = {\rm det} (g)\, {\rm det} (A)^2 
= {\rm det} (g) .$ Since ${\rm det} (g) = -1$, we obtain
\begin{equation} \label{detA}
 {\rm det} (A) = \pm 1 . 
\end{equation} 
One distinguishes two classes of transformations. {\it Proper}
Lorentz transformations are continuously connected with the
identity transformation $A = {\bf 1}$. All other Lorentz
transformations are called {\it improper}.
Proper transformations have necessarily ${\rm det} (A) = 1$. 
To have an improper Lorentz transformations it is sufficient, 
but not necessary, to have ${\rm det}(A)=-1$. For instance
$A=-{\bf 1}$ (space and time inversion) is an improper
Lorentz transformation with ${\rm det}(A)=+1$. 

Next the number of parameters, needed to identify a transformation 
in the group, follows from (\ref{Lg1}).  Since $A$ and $g$ are 
$4\times 4$ matrices, we have 16 equations for $4^2=16$ elements 
of $A$. But they are not all independent because of symmetry under 
transposition.  The off-diagonal equations are identical in pairs. 
Therefore, we have $4+6=10$ linearly independent equations for the 
16 elements of $A$. This leaves {\it six free parameters}, i.e., 
the Lorentz group is a six--parameter group. 

In the 19th century \index{} {\it Lie} invented the subsequent
procedure to handle these parameters. Let us now consider only 
proper Lorentz transformations. To construct $A$ explicitly, 
Lie made the ansatz
$$ A = e^L = \sum_{n=0}^{\infty} {L^n\over n!} , $$
where $L$ is a $4\times 4$ matrix. The determinant of $A$ is 
\begin{equation}
{\rm det} (A) = {\rm det} (e^L) = e^{{\rm Tr}(L)} . \label{det}
\end{equation}
Note that ${\rm det}(A)=+1$ implies that $L$ is traceless.
Equation (\ref{Lg1}) can be written
\begin{equation}
g \tilde{A} g = A^{-1} . \label{Lg2}
\end{equation}
From the definition of $L$, $\tilde{L}$ and the fact that 
$g^2 = {\bf 1}$ we have (note $(g\tilde{L}g)^n=g \tilde{L}^n g$ and 
${\bf 1} = 
(\sum_{n=0}^{\infty} L^n/n!)\, (\sum_{n=0}^{\infty} (-L)^n/n!)$) 
$$ \tilde{A}=e^{\tilde{L}},\ \ g\tilde{A}g=e^{g\tilde{L}g}
   ~~{\rm and}~~  A^{-1}=e^{-L}\ . $$
Therefore, (\ref{Lg2}) is equivalent to
$$ g\tilde{L}g=-L ~~{\rm or}~~ \widetilde{(gL)}=-gL . $$
The matrix $gL$ is thus antisymmetric and it is left as an 
exercise to show that the general form of $L$ is:
\begin{equation} \label{matrix_L}
   L = \left( \matrix{ 
    0      &  l^0_{\ 1} &  l^0_{\ 2} &  l^0_{\ 3} \cr
 l^0_{\ 1} &      0     &  l^1_{\ 2} &  l^1_{\ 3} \cr
 l^0_{\ 2} & -l^1_{\ 2} &     0      &  l^2_{\ 3} \cr
 l^0_{\ 3} & -l^1_{\ 3} & -l^2_{\ 3} &     0      \cr} \right).
\end{equation}
It is customary to expand $L$ in terms of six {\it generators}:
\begin{equation}
L=-\sum_{i=1}^3 (\phi_i S_i + \zeta_i K_i) ~~{\rm and}~~
A=e^{-\sum_{i=1}^3(\phi_iS_i+\zeta_i K_i)} . \label{Lie}
\end{equation}
The $S_i$ and $K_i$ matrices are defined by
\begin{equation} \label{rotation_matrices}
   S_1 = \left( \matrix{ 0 & 0 & 0 & 0 \cr
                         0 & 0 & 0 & 0 \cr
                         0 & 0 & 0 &-1 \cr
                         0 & 0 & 1 & 0 \cr} \right),\ \
   S_2 = \left( \matrix{ 0 & 0 & 0 & 0 \cr
                         0 & 0 & 0 & 1 \cr
                         0 & 0 & 0 & 0\cr
                         0 &-1 & 0 & 0 \cr} \right),\ \
   S_3 = \left( \matrix{ 0 & 0 & 0 & 0 \cr
                         0 & 0 &-1 & 0 \cr
                         0 & 1 & 0 & 0\cr
                         0 & 0 & 0 & 0 \cr} \right),
\end{equation}
and
\begin{equation} \label{boost_matrices}
  K_1 = \left( \matrix{ 0 & 1 & 0 & 0 \cr
                         1 & 0 & 0 & 0 \cr
                         0 & 0 & 0 & 0 \cr
                         0 & 0 & 0 & 0 \cr} \right),\ \
   K_2 = \left( \matrix{ 0 & 0 & 1 & 0 \cr
                         0 & 0 & 0 & 0 \cr
                         1 & 0 & 0 & 0 \cr
                         0 & 0 & 0 & 0 \cr} \right),\ \
   K_3 = \left( \matrix{ 0 & 0 & 0 & 1 \cr
                         0 & 0 & 0 & 0 \cr
                         0 & 0 & 0 & 0 \cr
                         1 & 0 & 0 & 0 \cr} \right) , 
\end{equation}
They satisfy the following \index{}{\it Lie algebra} commutation 
relations:
$$ [S_i,S_j] = \sum_{k=1}^3 \epsilon^{ijk} S_k,\ \
   [S_i,K_j] = \sum_{k=1}^3 \epsilon^{ijk} K_k,\ \
   [K_i,K_j] =-\sum_{k=1}^3 \epsilon^{ijk} S_k, $$
where the commutator of two matrices is defined by $[A,B]=AB-BA$
and $\epsilon^{ijk}$ is the completely antisymmetric Levi--Cevita 
tensor. Its definition in $n$--dimensions is
\begin{equation} \label{epsilon}
 \epsilon^{i_1 i_2 ... i_n} = \cases{
+1 ~~{\rm for}~~ (i_1, i_2, ..., i_n) 
~~{\rm being\ an\ even\ permutation\ of} ~~(1,2,...,n), \cr
-1 ~~{\rm for}~~ (i_1, i_2, ..., i_n) 
~~{\rm being\ an\ odd\ permutation\ of} ~~(1,2,...,n), \cr
 0 ~~{\rm otherwise}. } 
\end{equation}
To get the physical interpretation of equation (\ref{Lie}) for 
$A$, it is suitable to work out simple examples. First, let
$\vec{\zeta}=\phi_1=\phi_2=0$ and $\phi_3=\phi$. Then (left 
as exercise)
\begin{equation} \label{S3rot}
A = e^{-\phi\, S_3} =
   \left( \matrix{ 1 &     0      &     0      & 0 \cr
                   0 & \cos\phi & \sin\phi & 0 \cr
                   0 &-\sin\phi & \cos\phi & 0 \cr
                   0 &     0      &     0      & 1 \cr} \right) ,
\end{equation}
which describes a rotation by the {\it angle} $\phi$ (in the 
anti-clockwise sense) around the $\hat{e}_3$ axis. Next, let
$\vec{\phi}=\zeta_2=\zeta_3=0$ and $\zeta_1=\zeta$. Then
\begin{equation} \label{K1bost}
A = e^{-\zeta\, K_1} =
   \left( \matrix{ \cosh\zeta & -\sinh\zeta & 0 & 0 \cr
                  -\sinh\zeta &  \cosh\zeta & 0 & 0 \cr
                        0     &       0     & 1 & 0 \cr
0     &       0     & 0 & 1 \cr} \right) \label{mboost}
\end{equation}
is obtained, where $\zeta$ is known as the {\it boost parameter}
or {\it rapidity}. The structure is reminiscent to a rotation,
but with hyperbolic functions instead of circular, basically 
because of the relative negative sign between the space and 
time terms in eqn.(\ref{ILo1}). ``Rotations'' in the $x^0-x^i$ 
planes are boosts and governed by an hyperbolic gemometry, 
whereas rotations in the $x^i-x^j$ ($i\ne j$) planes are
governed by the ordinary Euclidean geometry.

Finally, note that the parameters
$\phi_i$, $\zeta_i$, $(i=1,2,3)$ turn out to be real, as
equation (\ref{Ltrans}) implies that the elements of $A$ have to
be real. In the next subsection relativistic kinematics is
discussed in more details.

\subsection{Basic relativistic kinematics}

The matrix (\ref{mboost}) gives the Lorentz boost transformation, which 
we discussed before in the 2D context (\ref{2DLT1}), (\ref{2DLT2}),
\begin{eqnarray}
  x'^{\, 0} &=&~~x^0 \cosh (\zeta ) - x^1 \sinh (\zeta )\ =\
                \gamma\,(x^0  - \beta\,x^1)\,, \label{LT1} \\
  x'^{\, 1} &=& -x^0 \sinh (\zeta ) + x^1 \cosh (\zeta )\ =\
                \gamma\,(x^1  - \beta\,x^0)\,, \label{LT2} \\
  x'^{\, i} &=& x^i,\ (i=2,3)\,. \label{LT3} 
\end{eqnarray}
To find the transformation law of an arbitrary vector $\vec{A}$ for 
a general relative velocity $\vec{v}$, it is convenient to decompose 
$\vec{A}$ into components parallel and perpendicular to $\vec{\beta} 
= \vec{v}/c$. Let $\hat{\beta}$ be the unit vector in $\vec{\beta}$ 
direction, 
$$ \vec{A} = A^{\|} \hat{\beta} + \vec{A}^{\perp} 
   ~~{\rm with}~~ A^{\|} = \hat{\beta} \vec{A}\ . $$
Then the Lorentz transformation law is
\begin{equation}
 A'^{\, 0} = A^0 \cosh (\zeta) - A^{\|} \sinh (\zeta)
      = \gamma (~~A^0 - \beta A^{\|}) , \label{VT1}
\end{equation}
\begin{equation}
 A'^{\, \|} = - A^0 \sinh (\zeta) + A^{\|} \cosh (\zeta)
        = \gamma (- \beta A^0 + A^{\|}) , \label{VT2}
\end{equation}
\begin{equation}
 \vec{A}'^{\, \perp} = \vec{A}^{\perp} . \label{VT3}
\end{equation}
Here I have reserved the subscript notation $A_{\|}$ and 
$\vec{A}_{\perp}$ for use in connection with covariant vectors: 
$A_{\|}=-A^{\|}$ and $\vec{A}_{\perp} = -\vec{A}^{\perp}$.
We proceed deriving the addition theorem for velocity vectors.

A particle moves with respect to $K'$ with velocity $\vec{u}^{\, '}$:
$$ x'^{\,i} = c^{-1} u'^{\,i} x'{\,^0}\,. $$
What is its velocity $\vec{u}$ with respect to $K$? Let us first 
assume that the velocity $\vec{v}$ between the frames is in $\hat{x}_1$ 
direction and rederive (\ref{velocity_add}). Substituting (\ref{LT2}) 
for $x'^{\,1}$ and (\ref{LT1}) for $x'^{\,0}$ gives
$$ \gamma\, (x^1-\beta x^0) = 
   c^{-1} u'^{\, 1} \gamma\, (x^0-\beta x^1)\,.$$
Sorting with respect to $x^1$ and $x^0$ yields
$$ \gamma\, \left(1+{u'^{\, 1} v\over c^2}\right)\, x^1 
    = c^{-1}\gamma\, (u'^1 + v)\, x^0 $$
and using the definition of the velocity in $K$, 
$\vec{x}=c^{-1} \vec{u}\, x^0$, one finds
\begin{equation} \label{u1pv}
 u^1 = c\, {x^1\over x^0} 
     = {u'^1 + v \over 1 + {u'^1 v / c^2}}\,. 
\end{equation}
Along similar lines we have for $i=1,2$:
$$ x^i = c^{-1}u'^{\,i}\,x^0 = c^{-1}u'^{\,i}\gamma\,(x^0-\beta\,x^1)\,. $$
Dividing by $x^0$ gives
$$ u^i = u'^{\,i}\gamma\,(1-v\,u^1/c^2) = u'^{\,i}\gamma\,
   \frac{1+v\,u'^{\,1}/c^2-v\,u'^{\,1}/c^2-v^2/c^2}{1+v\,u'^{\,1}/c^2}
   = u'^{\,i}\,\frac{\gamma\,(1-\beta^2)}{1+v\,u'^{\,1}/c^2}\,. $$
Using $1-\beta^2=1/\gamma^2$ (\ref{gamma}) we obtain for the $i=2,3$
components
\begin{equation} \label{uipv}
 u^i = {u'^i \over \gamma \left( 1+{u'^1 v / c^2}\right)},
        ~~(i=2,3). 
\end{equation}
To derive these equations, $\vec{v}$ was chosen along to the
$x^1$-axis. For general $\vec{v}$ one only has to decompose
$\vec{u}$ into its components parallel and perpendicular to
the $\vec{v}$
$$ \vec{u} = u^{\|} \hat{v} + \vec{u}^{\perp}\, , $$
where $\hat{v}$ is the unit vector in $\vec{v}$ direction,
and obtains
\begin{equation}
u^{\|} = {u'^{\, \|}+v \over 1 + {u'^{\, \|} v / c^2}} 
 ~~~{\rm and}~~~
\vec{u}^{\perp} = {\vec{u}^{\, '\, \perp} \over \gamma\, 
\left( 1 + {u'^{\, \|} v / c^2} \right) } . \label{addv}
\end{equation}
From this addition theorem of velocities it is obvious that the 
velocity itself is not part of of a 4-vector. The relativistic 
generalization is given in subsection~(\ref{dynamics}). It is
left as an exercise to relate these equations to the addition
theorem for the rapidity~(\ref{rapidity_add}.

The concepts of {\it world lines} in Minkowski space and {\it proper 
time} ({\it eigenzeit}) generalize immediately to 4D. Assume the 
particle moves with velocity $\vec{v}(t)$, then $d\vec{x}=\vec{\beta}
\,dx^0$ holds, and the infinitesimal invariant along its world line is
\begin{equation} \label{world_line}
  (ds)^2 = (dx^0)^2-(d\vec{x})^2 = (c\,dt)^2\,(1-\beta^2)\,.
\end{equation} 
Relations (\ref{dtau}) and (\ref{clocktrip}) hold as in 2D.

\subsection{Plane waves and the relativistic Doppler effect
            \label{sec_Doppler}}

Let us choose coordinates with respect to an inertial frame $K$.
In complex notation a plane wave is defined by the equation
\begin{equation} \label{wave}
W (x) = W(x^0,\vec{x}) = W_0\,
\exp [\, i\, (k^0\, x^0 - \vec{k}\cdot\vec{x} )\, ]\, ,
\end{equation}
where $W_0 = U_0 + i\, V_0$ is a complex {\it amplitude}. The vector 
$\vec{k}$ is called {\it wave vector} and $k^0$ is given by
\begin{equation} \label{wave_k0} 
  k^0 = \omega / c\ ,
\end{equation}
where $\omega$ is the {\it angular frequency} of the wave. 
Waves of this form may either propagate in a medium (water, air, 
shock waves, etc.) or in vacuum (light waves, particle waves in 
quantum mechanics). We are interested in the latter case, where 
$k^0$ and $\vec{k}$ combine to a relativistic 4-vector. In the 
other one a preferred inertial frame is defined, where the medium 
is at rest. 

The {\it phase} of the wave is defined by
\begin{equation} \label{phase}
  \Phi (x) = \Phi (x^0,\vec{x}) = k^0\, x^0 - \vec{k}\cdot\vec{x} =
  \omega\, t - \vec{k}\cdot\vec{x}\, . 
\end{equation}
When $(k^{\alpha})$ is a 4-vector, it follows that the phase is
a scalar, {\it invariant} under Lorentz transformations 
\begin{equation} \label{phaseinv}
  \Phi' (x') = k'_{\, \alpha}\, x'^{\, \alpha}
  = k_{\, \alpha}\, x^{\, \alpha} = \Phi (x)\, .
\end{equation}
That this is correct can be seen as follows:
For an observer at a fixed position $\vec{x}$ (note the term
$\vec{k}\cdot\vec{x}$ is then constant) the wave performs a 
periodic motion with {\it period} 
\begin{equation} \label{period}
 T = {2 \pi \over \omega} = {1 \over \nu}\, ,
\end{equation}
where $\nu$ is the {\it frequency}. In particular, the phase (and hence 
the wave) takes identical values on the two-dimensional hyperplanes 
perpendicular to $\vec{k}$. Let $\hat{k}$ be the unit vector in 
$\vec{k}$ direction. Decomposing $\vec{x}$ into components parallel 
and perpendicular to $\vec{k}$,
$\vec{x} = x^{\|}\, \hat{k} + \vec{x}^{\, \perp}$, the phase
becomes
\begin{equation} \label{phase2}
 \Phi = \omega\, t - k\, x^{\|}\, ,
\end{equation}
where $k=|\vec{k}|$ is the length of the vector $\vec{k}$. Phases
which differ by multiples of $2\pi$ give the same values for the
wave $W$. For example, when we take $V_0=0$, the real part of
the wave becomes
% \begin{equation} \label{wcos}
$$ W_x = U_0\, \cos (\omega\, t - k\, x^{\|}) $$
% \end{equation}
and $\Phi = 0$, $2\pi\,n$, $n=\pm 1, \pm 2, ...$ describes the 
wave crests. From (\ref{phase2}) it follows that the crests pass 
our observer with speed $\vec{u} =u\, \hat{k}$, where
\begin{equation} \label{wspeed}
 u = {\omega \over k} ~~{\rm as\ for}\ \Phi=0\ {\rm we\
 have}\ ~~x^{\|} = {\omega \over k}\, t\, .
\end{equation}
Let our observer count the number of wave crests passing by. How has 
the wave (\ref{wave}) to be described in another inertial frame $K'$?  
An observer in $K'$ counting the number of wave crests, passing through 
the same space-time point at which our first observer counts, must get 
the same number. The coordinates are just labels and the physics is the 
same in all systems. When in frame $K$ the wave takes its maximum at 
the space-time point $(x^{\alpha})$ it must also be 
at its maximum in $K'$ at the same space-time point in appropriately 
transformed coordinates $(x'^{\alpha})$. More generally, this holds 
for every value of the phase, because it is a scalar.

As $(k^{\alpha})$ is a 4-vector 
the transformation law for angular frequency and wave vector is
just a special case of equations (\ref{VT1}), (\ref{VT2}) and
(\ref{VT3})
\begin{equation} \label{Dop1}
 k'^{\, 0} = k^0 \cosh (\zeta) - k^{\|} \sinh (\zeta)
 = \gamma (k^0 - \beta k^{\|})\, ,
\end{equation}
\begin{equation}
 k'^{\, \|} = - k^0 \sinh (\zeta) + k^{\|} \cosh (\zeta)
 = \gamma (k^{\|} - \beta k^0 )\, ,
\end{equation}
\begin{equation}
 \vec{k}'^{\, \perp} = \vec{k}^{\perp}\, ,
\end{equation}
where the notation $k^{\|}$ and $k^{\perp}$ is with respect to 
the relative velocity of the two frames, $\vec{v}$. These
transformation
equations for the frequency and the wave vector describe the 
relativistic Doppler effect. To illustrate their meaning, we
consider a light source, which is emitted in $K$ and an observer 
in $K'$ moving in wave vector direction away from the source, i.e., 
$\vec{v}~\|~\vec{k}$. The equation for the wave speed~(\ref{wspeed}) 
implies 
$$ c ={\omega \over k}\ \Rightarrow\ k = |\vec{k}| =
      {\omega \over c} = k^0 $$
and chosing directions so that $k^{\,\|}=k$ holds, (\ref{Dop1}) becomes
$$ k'^{\, 0} = \gamma\, (k^0 - \beta\, k) = \gamma\, 
   (1-\beta)\, k^0 = k^0\, \sqrt{1-\beta \over 1+\beta} $$
or
$$ \omega' = {\nu' \over 2\pi} =  \omega\, \sqrt{1-\beta \over 1+\beta}
 =  {\nu \over 2\pi}\, \sqrt{1-\beta \over 1+\beta}\, . $$
Now, $c = \nu \lambda = \nu' \lambda'$, where $\lambda$ is the
wavelength in $K$ and $\lambda'$ the wavelength in $K'$. 
Consequently, we have
$$ \lambda' = \lambda\, \sqrt{1+\beta \over 1-\beta}\, . $$
For a receeding observer, or source receeding from the observer, 
$\beta>0$ in our conventions for $K$ and $K'$, and the wave length 
$\lambda'$ is larger than it is for a source at rest. This is an 
example of the {\it red-shift}, which is, for instance, of major 
importance when one analyzes spectral lines in astrophysics. Using 
the method of section~1, a single light signal suffices now to obtain 
position and speed of a distant mirror.

\subsection{Relativistic dynamics \label{dynamics}}

This section deals with the relativistic generalization of energy, 
momentum and their conservation laws. So far we have introduced two 
units, meter to measure distances and seconds to measure time. Both 
are related through a fundamental constant, the speed of light, so
that there is really only one independent unit up to now. In the
definition of the momentum a new, independent dimensional quantity 
enters, the mass of a particle. This unit is defined through the
gravitational law, which is out of the scope of these notes. 
Ideally, one would like to define the mass of a body as multiples
of the mass of an elementary particle, say an electron or proton.
This has remained too inaccurate. The mass unit has resisted 
modernization and is still defined through a
$$ 1~kilogram~[kg]\ =\ 1000~gram~[g] $$
standard object, a cylinder of platinum alloy which is kept at the 
International Bureau of Weights and Measures at S\'evres, France.

Let us consider a point-like particle in its rest-frame and denote 
its mass there by $m_0$. In any other frame the rest-mass of the 
particle is still $m_0$, which in this way is defined as a scalar.
It may be noted that most books in particle and nuclear physics
simply use $m$ to denote the rest-mass, whereas some books on special 
relativity employ the notation $m=\gamma m_0$ for a mass which
is proportional to the energy, i.e., the zero component of the 
energy-momentum vector introduced below. To avoid confusion, we 
use $m_0$ for the rest mass.

In the non-relativistic limit the momentum is defined by $\vec{p} = 
m_0 \vec{u}$. We want to define $\vec{p}$ as part of a relativistic 
4-vector $(p^{\alpha})$. Consider a particle at rest in frame $K$, 
i.e., $\vec{p}=0$. Assume now that frame $K'$ is moving with a small 
velocity $\vec{v}$ with respect to $K$. Then the non-relativistic limit 
is correct, and $\vec{p}{\, '} = -m_0\vec{v}$ has to hold approximately. 
On the other hand, the transformation laws (\ref{VT1}), (\ref{VT2}) and
(\ref{VT3}) for vectors (note $\vec{p}\,\|\,\vec{\beta}=\vec{v}/c$) 
imply
$$\vec{p}^{\ '} = \gamma\, (\vec{p} - \vec{\beta}\, p^0)\,.$$
For $\vec{p}=0$ we find $\vec{p}^{\ '}=-\gamma\,\vec{\beta}\,p^0$. 
As in the nonrelativstic limit $\gamma\,\beta\to\beta$, consistency 
requires $p^0=c\,m_0$ in the rest frame, so that we get $\vec{p}^{\ '}
=-m_0\,\gamma\,v$. Consequently, for a particle moving with velocity 
$\vec{u}$ in frame $K$ 
\begin{equation}
  \vec{p} = m_0\,\gamma\,\vec{u}
\end{equation}
is the relation between relativistic momentum and velocity. Due to 
the invariance of the scalar product $p_{\alpha}p^{\alpha}=(p^0)^2
-\vec{p}^{\, 2}=p'_{\alpha}p'^{\alpha}=m_0^2c^2$ holds and
\begin{equation} \label{p0}
 p^{\, 0} = +\sqrt{c^2\,m_0^2 + \vec{p}^{\ 2}} 
\end{equation} 
follows, which is of course consistent with calculating $p^0$ via 
the Lorentz transformation law (\ref{VT1}). As $c\,p^0$ has the 
dimension of energy, the relativistic energy of a particle is
\begin{equation} \label{renergy}
 E = c\, p^0=+\sqrt{c^4\,m_0^2+c^2\,\vec{p}^{\ 2}} =
  c^2\,m_0 + {\vec{p}{\, ^2} \over 2m_0} +\, ...\ ,
\end{equation} 
where the second term is just the non-relativistic kinetic energy 
$T=\vec{p}^{\ 2}/(2m_0)$. 
The first term shows that (rest) mass and energy can be transformed 
into one another~\cite{Ei1906}. In processes where the mass is 
conserved we just do not notice it. Using the mass definition of
special relativity books like \cite{Rind1}, $m=c\, p^0$, together 
with (\ref{renergy}) we obtain at this point the famous equation 
$E=m\, c^2$. Avoiding this definition of $m$, because it is not the
mass found in particle tables, where the mass of a particle is an
invariant scalar, the essence of Einstein's equation is captured by
$$ E_0 = m_0\, c^2\, , $$
where $E_0$ is the energy of a massive body (or particle) in its rest 
frame. 

Non-relativistic momentum conservation $\vec{p}_1+\vec{p}_2 =
\vec{q}_1+\vec{q}_2$, where $\vec{p}_i$, $(i=1,2)$ are the
momenta of two incoming, and $\vec{q}_i$, $(i=1,2)$ are the
momenta of two outgoing particles, becomes relativistic
{\it energy--momentum conservation}:
\begin{equation} \label{emom}
 p^{\alpha}_1 + p^{\alpha}_2 = q^{\alpha}_1 + q^{\alpha}_2 .
\end{equation} 
Useful formulas in relativistic dynamics are
\begin{equation} \label{gambet}
 \gamma = {p_0 \over m_0\, c} = {E \over m_0\, c^2}
  ~~~{\rm and}~~~ \beta = {|\vec{p}| \over p^0} . 
\end{equation} 
Further, the contravariant generalization of the velocity vector is 
given by
\begin{equation} \label{rvelocity}
 U^{\alpha} = {dx^{\alpha} \over d\tau} =
 \gamma u^{\alpha} ~~{\rm with}~~ u^0=c\, ,
\end{equation}
compare the definition of the infinitesimal proper time~(\ref{dtau}).
The relativistic generalization of the force is then the 4-vector
\begin{equation} \label{rforce}
 f^{\alpha} = {d\, p^{\alpha} \over d\, \tau} =
 m_0\, {d U^{\alpha}\over d \tau}\, ,
\end{equation}
where the last equality can only be used for particle with non-zero
rest mass.

% \subsection{Problems}
% \begin{description}
% \item{1.} Show 
% \end{description}

% \vfil\eject

\section{Maxwell Equations}

As before all considerations are in vacuum, as for fields in a
medium a preferred reference system exists. 
Maxwell's equations in their standard form in vacuum are
\begin{equation} \label{IMS0}
  \nabla\cdot\vec{E} = 4\pi \rho , ~~~\nabla \times \vec{B} -
  {1\over c} {\partial\vec{E}\over\partial t}={4\pi\over c}{\vec J}\ ,
\end{equation}
and
\begin{equation} \label{HMS0}
  \nabla\cdot\vec{B} = 0 , ~~~\nabla \times \vec{E} +
  {1\over c} {\partial \vec{B}\over \partial t} = 0\ .
\end{equation}
Here $\nabla$ is the Nabla operator, $\vec{E}$ is the {\it electric 
field} and $\vec{B}$ the {\it magnetic field} in vacuum. Equations 
(\ref{IMS0}) are the {\it inhomogeneous} and equations (\ref{HMS0}) 
are the {\it homogeneous} Maxwell equations in vacuum. When matter 
gets involved one introduces the {\it applied electric field} $\vec{D}$ 
and the {\it applied magnetic field} $\vec{H}$. Here we follow the 
convention 
of, for instance, Tipler~\cite{Tipler} and use the notation magnetic 
field for the measured field $\vec{B}$, in precisely the same way as it 
is done for the electric field $\vec{E}$, It should be noted that this 
is at odds with the notation in the book by Jackson~\cite{Jackson}, 
where (historically correct, but quite confusingly) $\vec{H}$ is called 
magnetic field and $\vec{B}$ magnetic flux or magnetic induction.

The charge density $\rho$ (charge per unit volume) and the current
density $\vec{J}$ (charge passing through a unit area per time unit)
are obviously given once a charge unit is defined through some 
measurement prescription.  From a theoretical point of view the 
electrical charge unit is best defined by the magnitude of the charge 
of a single electron (fundamental charge unit). In more conventional
units this reads
\begin{equation} \label{qelectron}
 q_e = 4.803 204 20 (19) \times 10^{-10}\, [esu] =
         1.602 176 462 (63)\times 10^{-19}\, Coulomb~[C] 
\end{equation}
where the errors are given in parenthesis. Definitions of the electric 
charge through measurement prescriptions rely presently on the current 
unit Amp\'ere
$[A]$ and are given in elementary physics textbooks like~\cite{Tipler}.
The numbers of (\ref{qelectron}) are from 1998~\cite{Co98}. The 
website of the National Institute of Standards and Technology (NIST) 
is given in this reference. Consult it for up to date information.

The choice of constants in the inhomogeneous Maxwell equations
defines units for the electric and magnetic field. The given
conventions $4\pi \rho$ and $(4\pi/c)\vec{J}$ are customarily
used in connection with Gaussian units, where the charge is
defined in electrostatic units $(esu)$.

In the next subsections the concepts of fields and currents are 
discussed in the relativistic context and the electromagnetic field 
equations follow in the last subsection.

\subsection{Fields and currents}

A tensor field is a tensor function which depends on the
coordinates of Minkowski space: 
$$ T^{...\alpha ...\beta ...} = 
   T^{...\alpha ...\beta ...} (x)\, . $$
It is called {\it static} when there is no time dependence. For
instance $\vec{E} (\vec{x})$ in electrostatics would be a static
vector field in three dimensions. We are here, of course, primarily 
interested in contravariant or covariant fields in four dimensions, 
like vector fields $A^{\alpha} (x)$.

Suppose $n$ electric charge units are contained in a small volume $v$,
so that we can talk about the position $\vec{x}$ of this volume. The 
corresponding electrical charge density at the position of that
volume is then just $\rho=n/v$ and the electrical current is defined
as the charge that passes per unit time through a surface element of
such a volume. We demand now that the electric charge density
$\rho$ and the electric current $\vec{J}$ form a 4-vector:
$$ (J^{\alpha}) = \left( \matrix{ c\rho \cr \vec{J}\cr} \right)\, .$$
The factor $c$ is introduced by dimensional reasons and
we have suppressed the space-time dependence, i.e.,
$J^{\alpha} = J^{\alpha} (x)$ forms a vector field. It is left as
an exercise to write down the 4-current for a point particle of 
elementary charge $q_e$.

The {\it continuity equation} takes the simple invariant form
\begin{equation} \label{continuityeqn}
 \partial_{\alpha} J^{\alpha} = 0\, .
\end{equation}
The charge of a point particle in its rest frame is also an invariant:
$$ c^2 q_0^2 = J_{\alpha} J^{\alpha}\, .$$

\subsection{The inhomogeneous Maxwell equations}

The inhomogeneous Maxwell equations are obtained by writing down the
simplest covariant equation which yields a 4-vector as first order
derivatives of six fields. From undergraduate E\&M we remember the 
electric and magnetic fields, $\vec E$ and $\vec B$, with together
six free parameters. We now like to describe them in covariant form. 
A 4-vector is unsuitable as we like to describe six quantities $E^x, 
E^y, E^z$ and $B^x, B^y, B^z$. Next, we may try a rank two tensor 
$F^{\alpha\beta}$. Then we have $4\times 4 =16$ quantities at our 
disposal, i.e., too many. But, one may observe that a symmetric 
tensor stays symmetric under Lorentz transformation and an 
antisymmetric tensor stays antisymmetric.
Hence, instead of looking at the full second rank tensor one has to 
consider its symmetric and antisymmetric parts separately.

By requesting $F^{\alpha\beta}$ to be antisymmetric, 
\begin{equation}
 F^{\alpha\beta} = - F^{\beta\alpha} , \label{Fab}
\end{equation}
the number of independent parameters is  reduced to precisely six. The 
diagonal elements do now vanish,
$$ F^{00} = F^{11} = F^{22} = F^{33} = 0 $$
and other elements follow from $F^{\alpha\beta}$ with $\alpha < \beta$ 
through (\ref{Fab}). As desired, this gives $(16-4)/2=6$ independent 
elements.

Up to an over--all factor, which is chosen by convention, the only way to 
obtain a 4-vector through differentiation of $F^{\alpha\beta}$ is 
\begin{equation} \label{IMEQ}
\partial_{\alpha}F^{\alpha\beta}={4\pi\over c} J^{\beta}\ .
\end{equation}
This is the {\it inhomogeneous Maxwell equation} in covariant
form. Note that it determines the physical dimensions of the electric 
fields, the factor $4\pi/c$ on the right-hand side corresponds to
Gaussian units. The continuity equation~(\ref{continuityeqn}) is a 
simple consequence of the inhomogeneous Maxwell equation
$$ {4\pi\over c} \partial_{\beta} J^{\beta} = 
\partial_{\beta} \partial_{\alpha} F^{\alpha\beta} = 0 $$
because the contraction of the symmetric tensor $(\partial_{\beta} 
\partial_{\alpha})$ with the antisymmetric tensor $F^{\alpha\beta}$ 
is zero.

To relate the elements of the $F^{\alpha\beta}$ tensor to the $\vec{E}$
and $\vec{B}$ fields, let us choose $\beta=0,1,2,3$ and compare 
equation~(\ref{IMEQ}) with the inhomogeneous Maxwell equations in 
their standard form~(\ref{IMS0}). For instance, $\partial_{\alpha} 
F^{\alpha\,0}=\nabla\vec{E}=4\pi\,\rho$ yields the $F^{i\,0}=E^i$, 
the first column of the $F^{\alpha\beta}$ tensor. Extending this 
comparision it the other $\beta$ values, the final result is
\begin{equation} \label{FEB}
   (F^{\alpha\beta}) = \left(
\matrix{ 0   & -E^x & -E^y & -E^z \cr
         E^x &  0   & -B^z &  B^y \cr
         E^y &  B^z &  0   & -B^x \cr
         E^z & -B^y &  B^x &  0   \cr} \right)\, .
\end{equation}
Or, in components
\begin{equation} \label{EMfields}
 F^{i0}=E^i ~~{\rm and}~~ F^{ij} = -\sum_k \epsilon^{ijk}\,B^k
 \Leftrightarrow B^k = - {1\over 2}\, \sum_i\sum_j 
 \epsilon^{kij} F^{ij}\ .
\end{equation}
Next, $F_{\alpha\beta}=g_{\alpha\gamma}g_{\beta\delta}F^{\gamma\delta}$
implies:
$$ F_{0i}=-F^{0i},\ F_{00}=F^{00}=0,\ F_{ii}=F^{ii}=0,\
 ~~~{\rm and}~~~ F_{ij} = F^{ij} . $$
Consequently,
\begin{equation}
   (F_{\alpha\beta}) = \left(
\matrix{ 0   &  E^x &  E^y &  E^z \cr
        -E^x &  0   & -B^z &  B^y \cr
        -E^y &  B^z &  0   & -B^x \cr
        -E^z & -B^y &  B^x &  0   \cr} \right).
\end{equation}

\subsection{Four-potential and homogeneous Maxwell equations}

We remember that the electromagnetic
fields may be written as derivatives of appropriate potentials. The only
covariant option are terms like $\partial^{\alpha} A^{\beta}$. To make
$F^{\alpha\beta}$ antisymmetric, we have to subtract 
$\partial^{\beta} A^{\alpha}$:
\begin{equation} \label{DaAb}
F^{\alpha\beta}=\partial^{\alpha}A^{\beta}-\partial^{\beta}A^{\alpha} .
\end{equation}
It is amazing to note that the homogeneous Maxwell equations follow now
for free. The dual electromagnetic tensor is defined by
\begin{equation}
  {^*F}^{\alpha\beta} = {1\over 2} \epsilon^{\alpha\beta\gamma\delta}
                        F_{\gamma\delta} , \label{DFab}
\end{equation}
and it holds
\begin{equation}
  \partial_{\alpha} {^*F}^{\alpha\beta} = 0 . \label{HMEQ}
\end{equation}
Proof:
$$ \partial_{\alpha} {^*F}^{\alpha\beta} = 
{1\over 2} \left( \epsilon^{\alpha\beta\gamma\delta} 
\partial_{\alpha} \partial_{\gamma} A_{\delta} - 
\epsilon^{\alpha\beta\gamma\delta} 
\partial_{\alpha} \partial_{\delta} A_{\gamma} \right) = 0 . $$

This first term is zero due to (\ref{SA}), because 
$\epsilon^{\alpha\beta\gamma\delta}$ is antisymmetric in $(\alpha,\gamma)$,
whereas the derivative $\partial_{\alpha}\partial_{\gamma}$ is symmetric
in $(\alpha,\gamma)$. Similarly the other term is zero. The homogeneous 
Maxwell equation is related to the fact that the right-hand side of 
equation (\ref{DaAb}) expresses six fields in terms of a single 4-vector.
An equivalent way to write (\ref{HMEQ}) is the equation
\begin{equation} \label{HMEQa}
    \partial^{\alpha}F^{\beta\gamma} + \partial^{\beta}F^{\gamma\alpha} 
  + \partial^{\gamma} F^{\alpha\beta} = 0\ .
\end{equation}
The proof is left as an exercise to the reader.
Let us mention that the homogeneous Maxwell equation (\ref{HMEQ}) or
(\ref{HMEQa}), and hence our demand that the field can be written in 
the form (\ref{DaAb}), excludes magnetic monopoles.

The elements of the dual tensor may be calculated from 
their definition (\ref{DFab}). For example, 
$${^*F}^{02} = \epsilon^{0213} F_{13} = -F_{13} = -B^y , $$
where the first step exploits the anti-symmetries 
$\epsilon^{0231}=-\epsilon^{0213} $ and $F_{31}=-F_{13}$. Calculating
six components, and exploiting antisymmetry of 
${^*F}^{\alpha\beta}$, we arrive at
\begin{equation}
   ({^*F}^{\alpha\beta}) = \left(
\matrix{ 0   & -B^x & -B^y & -B^z \cr
         B^x &  0   &  E^z & -E^y \cr
         B^y & -E^z &  0   &  E^x \cr
         B^z &  E^y & -E^x &  0   \cr} \right).
\end{equation}
The homogeneous Maxwell equations in their form (\ref{HMS0})
provide a non--trivial consistency check for (\ref{HMEQ}),
which is of course passed. 

A notable observation is that equation (\ref{DaAb}) does not determine
the potential uniquely. Under the transformation
\begin{equation}
A^{\alpha} \mapsto A'^{\alpha} = 
A^{\alpha} + \partial^{\alpha} \psi , \label{gauge}
\end{equation}
where $\psi = \psi (x)$ is an arbitrary scalar function, the
electromagnetic field tensor is invariant: 
$F'^{\alpha\beta} = F^{\alpha\beta}$, as follows immediately from
$\partial^{\alpha}\partial^{\beta}\psi - 
 \partial^{\beta}\partial^{\alpha}\psi = 0$. The transformations
(\ref{gauge}) are called {\it gauge transformation}$^{1}$. The 
choice of a convenient gauge is at the heart of many calculations.

\footnotetext[1]{In quantum field theory these are the gauge 
transformations of second kind. Gauge transformations of first kind 
transform fields by a constant phase, whereas for gauge transformation 
of the second kind a space--time dependent function is encountered.} 

\subsection{Lorentz transformation for the electric and magnetic fields}

The electric $\vec{E}$ and magnetic $\vec{B}$ fields are {\it not}
components of a Lorentz four-vector, but part of the rank two 
the electromagnetic field $(F^{\alpha\beta})$ given by (\ref{FEB}).
As for any Lorentz tensor, we immediately know the behavior of 
$(F^{\alpha\beta})$ under Lorentz transformations
\begin{equation}
  F'^{\alpha\beta} = a^{\alpha}_{\ \gamma}\,a^{\beta}_{\ \delta}\,
  F^{\gamma\delta}\ .
\end{equation}
Using the explicit form (\ref{mboost}) of $A=(a^{\alpha}_{\ \beta})$ 
for boosts in the $x^1$ direction and (\ref{FEB}) for the relation 
to $\vec{E}$ and $\vec{B}$ fields, it is left as a straightforward 
exercise to derive the transformation laws
\begin{equation} \label{LTE}
  \vec{E}^{\,'} = \gamma\,\left( \vec{E} + \vec{\beta} \times
  \vec{B} \right) - {\gamma^2\over \gamma +1}\,\vec{\beta}\,
  \left( \vec{\beta}\,\vec{E} \right)\ ,
\end{equation}
and
\begin{equation} \label{LTB}
  \vec{B}^{\,'} = \gamma\,\left( \vec{B} - \vec{\beta} \times
  \vec{E} \right) - {\gamma^2\over \gamma +1}\,\vec{\beta}\,
  \left( \vec{\beta}\,\vec{B} \right)\ .
\end{equation}


\subsection{Lorentz force}

Relativistic dynamics of a point particle (more generally any mass
distribution) gets related to the theory of electromagnetic fields,
because an electromagnetic field causes a change of the 4-momentum
of a charged particle. On a deeper level this phenomenon is related 
to the conservation of energy and momentum and the fact that an
electromagnetic field carries energy as well as momentum. Here we 
are content with finding the Lorentz covariant force.

We consider a charged point particle in an electromagnetic field
$F^{\alpha\beta}$. Here {\it external} means from sources other
than the point particle itself and that the influence of the point
particle on these other sources (possibly causing a change of the
field $F^{\alpha\beta}$) is neglected. The infinitesimal change of
the 4-momentum of a point point particle is $dp^{\alpha}$ and
assumed to be proportional to (i)~its charge $q$ and (ii)~the
external electromagnetic field $F^{\alpha\beta}$. This means, we
have to contract $F^{\alpha\beta}$ with some infinitesimal covariant 
vector to get $dp^{\alpha}$. The simplest choice is $dx_{\beta}$, what
means that the amount of 4-momentum change is proportional to the 
space-time length at which the particle experiences the electromagnetic 
field. Hence, we have determined $dp^{\alpha}$
up to a proportionality constant, which depends on the choice of
units. Gaussian units are defined by choosing $c^{-1}$ for this
proportionality constant and we have
\begin{equation}   \label{pm_dLF}
  dp^{\alpha} = \pm {q\over c}\, F^{\alpha\beta}\, dx_{\beta}\ .
\end{equation}
As discussed in the next section, it is a consequence of energy 
conservation, which is in this context known as {\it Lenz law}, 
that the force between charges of equal sign has to be repulsive. 
This corresponds to the plus sign and we arrive at
\begin{equation}   \label{dLF}
  dp^{\alpha} = {q\over c}\, F^{\alpha\beta}\, dx_{\beta}\ .
\end{equation}
Experimental measurements are of
course in agreement with this sign. The remarkable  point is that
energy conservation and the general structure of the theory already
imply that the force between charges of equal sign has to be
{\it repulsive}. Therefore, despite the similarity of the Coulomb's
inverse square force law with Newton's law it impossible
to build a theory of gravity along the lines of this chapter,
i.e., to use the 4-momentum $p^{\alpha}$
as source in the inhomogeneous equation~(\ref{IMEQ}). The
resulting force would necessarily be repulsive. Experiments
show also that positive and negative electric charges exist and
deeper insight about their origin comes from the relativistic Lagrange 
formulation, which includes Dirac's equation for electrons and leads 
to Quantum Electrodynamics.

Taking the derivative with respect to the proper time, we
obtain the 4-force acting on a charged particle, called
{\it Lorentz force},
\begin{equation}  \label{LF}
f^{\alpha} = {dp^{\alpha}\over d\tau} =
{q\over c} F^{\alpha\beta}U_{\beta} .
\end{equation}
As in equation~(\ref{rforce}) $f^{\alpha} = m_0\, d U^{\alpha}/d\tau$
holds for non-zero rest mass and the definition of the
contravariant velocity is given by equation~(\ref{rvelocity}). 

Using the representation~(\ref{FEB}) of the electromagnetic field
the time component of the relativistic Lorentz force, which describes 
the change in energy, is 
\begin{equation}  \label{LFtime}
 f^0 = {dp^0\over d\tau} =
 - {q\over c} \left( \vec{E} \vec{U} \right)\ .
\end{equation}
To get the space component of the Lorentz force we use (\ref{FEB}) 
and (\ref{EMfields}) and get the equality
$$ {q\over c}\sum_{j=1}^3F^{ij}U_j = 
-{q\over c}\sum_{j=1}^3\sum_{k=1}^3\epsilon^{ijk}B^kU_j $$
The space components combine into the well-known {\it Lorentz force}
\begin{equation} \label{LFspace}
\vec{f} = q\,\gamma\,\vec{E} 
	+ {q\over c}\, \vec{U} \times \vec{B}\,,
\end{equation}
which reveals that the relativistic velocity
(\ref{rvelocity}) of the charge $q$ and not its velocity $\vec{v}$
enters the force equation. This allows, for instance, correct 
force calculations for fast flying electrons in a magnetic field.
The equation (\ref{LFspace}) for $\vec{f}$ can be used to define 
a measurement prescription for an electric charge unit.
% \hfill\break % \vfill\eject

\section{Faraday's Law}

From the homogeneous Maxwell equations (\ref{HMEQ}) we have 
\begin{equation}
  \nabla \times \vec{E} + {1\over c}\,
  {\partial\vec{B} \over \partial t} = 0\, . 
\end{equation}
This equation is the \index{Faraday's law}{\it differential form of 
Faraday's law}: A changing magnetic field induces an electric field. 
In the following we derive the integral form, which is needed for 
circuits of macroscopic extensions. We integrate over a simply 
connected surface $S$ and use Stoke's theorem to convert the integral 
over $\nabla\times\vec{E}$ into a closed line integral along
the boundary $C$ of $S$:
$$ \int_S (\,\nabla\times\vec{E}\,)\cdot d\vec{a} =
  \oint_C \vec{E}\cdot d\vec{l} = - {1\over c} \int_S
  {\partial\vec{B} \over \partial t}\cdot d\vec{a}\, .$$
On the right-hand side we eliminate the partial derivative
${\partial/\partial t}$ using
$$ {d\ \over dt} = {\partial\ \over\partial t} +
  \vec{v}\cdot\nabla~~~\left( {\rm note}~~ \vec{v} =
  {\partial\vec{x} \over\partial t} =
  \sum_{i=1}^3 \hat{e}^i\,{\partial x^i\over\partial t}\right) $$
to get 
\begin{equation} \label{F1}
   \oint_C \vec{E}\cdot d\vec{l} =
 - {1\over c} {d\ \over dt} \int_S \vec{B}\cdot d\vec{a}
 + {1\over c} \int_S (\vec{v}\cdot\nabla)\,\vec{B}\cdot d\vec{a}
\end{equation}
Using the other homogeneous Maxwell equation, $\nabla\cdot\vec{B}=0$,
and that the $\partial/\partial x^i$ derivatives of $\vec{v}$ vanish
(e.g., $(\partial/\partial x^1)\,(\partial x^1/\partial t) = (\partial
/\partial t)\,(\partial x^1/\partial x_1) = 0$), the vector identity
$$[\nabla\times (\,\vec{a}\times \vec{b}\,)] = 
  (\,\nabla\cdot\vec{b}\,)\,\vec{a}+(\,\vec{b}\cdot\nabla\,)\,\vec{a} 
 -(\,\nabla\cdot\vec{a}\,)\,\vec{b}-(\,\vec{a}\cdot\nabla\,)\,\vec{b}\,,$$
gives
$$ \nabla\times \left( \vec{B} \times \vec{v} \right) =
   \vec{v}\,\left( \nabla\cdot\vec{B}\right) +
   \left( \vec{v}\cdot\nabla \right)\,\vec{B} =
   \left( \vec{v}\cdot\nabla \right)\,\vec{B} $$
and we transform the last integral in (\ref{F1}) as follows
\begin{eqnarray} \nonumber
  {1\over c} \int_S (\vec{v}\cdot\nabla)\, \vec{B}\cdot d\vec{a} &=&
  - {1\over c} \int_S \nabla\times \left(\vec{v}\times\vec{B}\right)
    \cdot d\vec{a}\, = \\ \nonumber
  - {1\over c} \oint_C \left(\vec{v}\times\vec{B}\right)\cdot d\vec{l}
  &=& - \oint_C \left(\vec{\beta}\times\vec{B}\right)\cdot d\vec{l}\,,
\end{eqnarray}
where Stoke's theorem has been used and $\vec{\beta}=\vec{v}/c$. We 
re-write equation~(\ref{F1}) with both encountered line integral on 
the left-hand side
\begin{equation} \label{F2}
  \oint_C \left(\vec{E}+\vec{\beta}\times\vec{B}\right)
  \cdot d\vec{l} = - {1\over c} {d\ \over dt} \Phi_m\,,
\end{equation}
where $\Phi$ is called \index{Magnetic flux}{\it magnetic flux} and 
defined by
\begin{equation} \label{magflux}
  \Phi_m = \int_S \vec{B}\cdot d\vec{a}\ .
\end{equation}
Equation (\ref{F2}) is the fully relativistic version of Faraday's law. 
The velocity $\vec{\beta}=\vec{v}/c$ in equation (\ref{F2}) refers to 
the velocity of the line element $d\vec{l}$ with respect to the inertial 
frame in which the calculation is done. In the frame co-moving with 
apparatus, normally the Lab frame, the velocity differences between
different line element sections are small so that we can neglect the 
$\vec{\beta} \times\vec{B}$ contribution:
\begin{equation} \label{Faraday}
  \epsilon_{\rm emf} = \oint_C \vec{E}\cdot d\vec{l}
  = - {1\over c} {d\ \over dt} \Phi_m\,,
\end{equation}
where $\epsilon_{\rm emf}$ is called\index{Electromotive force} 
{\it electromotive force} (emf)\index{emf}. In this approximation 
\index{Faraday's law}\index{Induction}{\it Faraday's Law of Induction}
is stated in most test books. Due to our initial treatment of special 
relativity we do not face the problem to work out its relativistic 
generalization, but instead obtained (\ref{Faraday}) as an approximation
of the generally correct law (\ref{F2}).

\subsection{Lenz's Law\label{Lenz_law}}

With the Lorentz force~(\ref{LF}) given, Energy conservation determines 
the minus sign on the right-hand side of Faraday's law~(\ref{Faraday}). 
This is known as \index{Lenz law}{\it Lenz law}. For closed, 
conducting circuits the emf~(\ref{Faraday}) will induce a current, 
whose magnitude depends on the resistance of the circuit. Lenz's law 
states: {\it The induced emf and induced current are in such a 
direction as to oppose the change that produces them.} Tipler 
\cite{Tipler} gives many examples. To illustrate the connection 
with energy conservation, we discuss one of them.

We consider a permanent bar magnet moving towards a closed loop that 
has a resistance $R$. The north pole of the bar magnet is defined so 
that the magnetic field points out of it. We arrange the north--south
axis of the magnet perpendicular to the surface spanned by the loop 
and move the magnet toward the loop. The magnetic field through the 
loop gets stronger when the magnet is approaching and a current is 
induced in the loop. The direction of the current is such that its 
magnetic field is opposite to that of the magnet. Effectively the 
loop becomes a magnet with north pole towards the bar magnet. The 
result is a {\it repulsive} force between bar magnet and loop. Work 
against this force is responsible for the induced current and its 
associated heat in the loop. Would the sign of the induced current 
be different, an attractive force would result and the resulting 
acceleration of the bar magnet as well as the heat in the loop 
would violate energy conservation. Note that pulling the bar magnet 
out of the loop does also produce energy.

In our treatment the sign of Faraday's law is already given by
the electromagnetic field equation and energy conservation 
determines the sign in equation (\ref{pm_dLF}) for the Lorenz
force.

\begin{thebibliography}{12}
\addcontentsline{toc}{chapter}{Bibliography}%

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\end{thebibliography}

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\end{document}
