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\begin{document}

\section{Kinematics of Special Relativity}

The special theory of relativity arose out of a conflict between the
ideas of mechanics as developed by Galileo and Newton, and the ideas
of electromagnetism.  For this reason relativity is often discussed
after electromagetism is developed.  However, the the connections
between relativity and electromagnetism are high-level and difficult
to understand the first time around.  Furthermore, special relativity
is actually a generally valid extension to Newtonian mechanics that is
needed when objects move at very high speeds, and is only
coincidentally related to electromagnetism.  For this reason I choose
to discuss relativity before electromagnetism.

The only fact from electromagnetism that we need is introduced now:
There is a maximum speed at which objects can travel.  This is
coincidentally equal to the speed of light in a vacuum, $c = 3 \times
10^8 \unit{m} \unit{s}^{-1}$.  Furthermore, a measurement of the speed
of a particular light beam yields the same answer regardless of the
speed of the light source or the speed at which the measuring
instrument is moving.

This rather bizarre experimental result is in contrast to what occurs
in classical mechanics.  If two cars pass a pedestrian standing on a
curb, one at $20 \unit{m} \unit{s}^{-1}$ and the other at $50 \unit{m}
\unit{s}^{-1}$, the faster car appears to be moving at $30 \unit{m}
\unit{s}^{-1}$ relative to the slower car.  However, if a light beam
moving at $3 \times 10^8 \unit{m} \unit{s}^{-1}$ passes an
interstellar spaceship moving at $2 \times 10^8 \unit{m}
\unit{s}^{-1}$, then the light beam appears to be moving at $3 \times
10^8 \unit{m} \unit{s}^{-1}$ to occupants of the spaceship, not $1
\times 10^8 \unit{m} \unit{s}^{-1}$.  Furthermore, if the spaceship
beams a light signal forward to its (stationary) destination planet,
then the resulting beam appears to be moving at $3 \times 10^8
\unit{m} \unit{s}^{-1}$ to instruments at the destination, not $5
\times 10^8 \unit{m} \unit{s}^{-1}$.

The fact that we are talking about light beams is only for
convenience.  Any other means of sending a signal at the maximum
allowed speed would result in the same behavior.  We therefore cannot
seek the answer to this apparent paradox in the special properties of
light.  Instead we have to look to the basic nature of space and time.

\subsection{Spacetime Thinking}

\begin{figure}
\begin{center}
\psfig{figure=spacetime.eps,height=3in,width=3in,angle=0}
\end{center}
\caption{Spacetime diagram showing an {\em event} and a {\em world
line}.}
\label{spacetime}
\end{figure}

In order to gain an understanding of special relativity it is
important to begin thinking of space and time as being different
dimensions of a four-dimensional space called spacetime.  Actually,
since we can't visualize four dimensions very well, it is easiest to
start with only one space dimension and the time dimension.  Figure
\ref{spacetime} shows a graph with time plotted on the vertical axis
and the one space dimension plotted on the horizontal axis.  An {\em
event} is something that occurs at a particular time and a particular
point in space.  (``Julius X. wrecks his car in Lemitar, NM on 21 June
at 6:17 PM.'')  A {\em world line} is a plot of the position of some
object as a function of time (more properly, the time of the object as
a function of position) on a spacetime diagram.  Thus, a world line is
really a line in spacetime, while an event is a point in spacetime.

In a spacetime diagram the slope of a world line has a special meaning.
Notice that a vertical world line means that the object it represents
does not move --- the velocity is zero.  If the object moves to the
right, then the world line tilts to the right, and the faster it moves,
the more the world line tilts.  Quantitatively, we say that
\begin{equation}
     velocity = \frac{1}{slope~of~world~line} .
\label{slope}
\end{equation}
Notice that this works for negative slopes and velocities as well as
positive ones.  If the object changes its velocity with time, then the
world line is curved, and the instantaneous velocity at any time is
the inverse of the slope of the tangent to the world line at that
time.

The hardest thing to realize about spacetime diagrams is that they
represent the past, present, and future all in one diagram.  Thus,
spacetime diagrams don't change with time --- the evolution of physical
systems is represented by looking at successive horizontal slices in
the diagram at successive times.  {\em Spacetime diagrams represent
evolution, but they don't evolve themselves.}

\begin{figure}
\begin{center}
\psfig{figure=spacetime2.eps,height=3in,width=3in,angle=0}
\end{center}
\caption{Scaled spacetime diagram showing world lines of light passing
left and right through the origin.}
\label{spacetime2}
\end{figure}

If time and position are simply different dimensions of the same
abstract space, then they should have the same units.  The easiest way
to arrange this is to multiply time by the maximum speed, $c$,
resulting in the kind of spacetime diagram shown in figure
\ref{spacetime2}.  Notice that world lines of light have slope $\pm 1$
when the time axis is scaled this way.  Furthermore, (\ref{slope}) must
be revised to read
\begin{equation}
     v = \frac{c}{slope} .
\label{slope2}
\end{equation}

\subsection{The Principle of Relativity}

The {\em principle of relativity} states that the laws of physics are
the same in all {\em inertial} reference frames.  An inertial
reference frame is one that is not accelerated.  Reference frames
attached to a car at rest and to a car moving steadily down the
freeway at $30 \unit{m} \unit{s}^{-1}$ are both inertial.  A reference
frame attached to a car accelerating away from a stop light is not
inertial.  Newton's second law, $F = ma$, works in both stationary and
moving inertial frames, but not in accelerated frames.  Newton's law
can be fixed up in non-inertial frames by adding an additional {\em
inertial force} equal to minus the mass times the acceleration of the
reference frame to the left side of $F = ma$.  However, this is just a
fudge factor which is invoked to preserve appearances --- the fact
remains that $F = ma$ in unmodified form doesn't work in an
accelerated reference frame.

The principle of relativity is not cast in stone, but is an educated
guess or hypothesis based on extensive experience.  If the principle
of relativity weren't true, we would have to do all our calculations
in some preferred reference frame in which we knew $F = ma$ to be
true.  This would be very annoying.  However, the more fundamental
problem is that we have no idea what the velocity of this preferred
frame might be.  Does it move with the earth?  That would be very
earth-centric.  How about the velocity of the center of our galaxy
or the mean velocity of all the galaxies?  Rather than face the issue
of a preferred reference frame, physicists have chosen to stick with
the principle of relativity.

Notice that the constancy of the speed of light in all reference
frames is consistent with the principle of relativity.  (Why???)
However, as noted above, it is inconsistent with our notions as to how
velocities add, or alternatively, how we think the world should look
from reference frames moving at different speeds.  The classical way
of understanding the view from different reference frames is called
``Galilean relativity''.  The new way that reconciles the behavior of
objects moving at very high speeds is called ``Einsteinian
relativity''.  Einstein's great contribution was to discover the laws
that tell us how the world looks from reference frames moving at high
speeds relative to each other.  These laws constitute a {\em geometry
of spacetime}, and from them all of special relativity can be derived.

\subsection{Postulates of Special Relativity}

\begin{figure}
\begin{center}
\psfig{figure=pythag.eps,height=3in,width=3in,angle=0}
\end{center}
\caption{Triangle for Pythagorean theorem in spacetime.}
\label{pythag}
\end{figure}

All of the observed facts about spacetime can be derived from two
postulates:

\begin{itemize}
\item
Spacetime obeys a modified Pythagorean theorem, which gives the
distance, $I$, in spacetime as
\begin{equation}
     I^2 = X^2 - c^2 T^2 ,
\label{pytheqn}
\end{equation}
where $X$, $T$, and $I$ are defined in figure \ref{pythag}.
\item
Whether two events appear to be simultaneous depends on the coordinate
system from which they are viewed.
\end{itemize}
Let us discuss these postulates in turn.

The Pythagorean theorem of spacetime differs from the usual
Pythagorean theorem in two ways.  First, the vertical side of the
triangle is multiplied by $c$.  This is a trivial scale factor that
gives time the same units as space.  Second, the right side of
(\ref{pytheqn}) has a minus sign rather than a plus sign.  This
highlights a fundamental difference between spacetime and the ordinary
$xyz$ space in which we live.  Spacetime is said to have a {\em
non-Euclidean} geometry --- in other words, the normal rules of
geometry that we learn in high school don't always work for spacetime!

The main consequence of the minus sign in (\ref{pytheqn}) is that
$I^2$ can be negative and therefore $I$ can be imaginary.
Furthermore, in the special case where $X = \pm cT$, we actually have
$I = 0$ even though $X,~T \ne 0$ --- i. e., the ``distance'' between
two well-separated events can be zero.  Clearly, spacetime has some
weird properties!

\begin{figure}
\begin{center}
\psfig{figure=simul.eps,height=3in,width=3in,angle=0}
\end{center}
\caption{Sketch of coordinate axes for a moving reference frame, $x' ,~
ct'$.  The meanings of the events A-E are discussed in the text.
The lines tilted at $\pm 45^{\circ}$ are the world lines of light
passing through the origin.}
\label{simul}
\end{figure}

The quantity $I$ is usually called an {\em interval} in spacetime.  If
the two events defining the end points of the interval have the same
$t$ value, then the interval is the ordinary space distance between
the two events.  On the other hand, if they have the same $x$ value,
then the interval is equal to $ic$ times the time interval between the
events, where $i = (-1)^{1/2}$.  This much is clear from figure
\ref{pythag}.  The interpretation of $I$ in the more general case must
await our discussion of the second postulate.  However, generally
speaking, if $I$ is real, the interval is called {\em spacelike},
while an imaginary interval is called {\em timelike}.

The classical way of thinking about {\em simultaneity} is so ingrained
in our everyday habits that we have a great deal of difficulty
adjusting to what special relativity has to say about this subject.
Indeed, understanding how relativity changes this concept is the
single most difficult part of the theory --- once you understand this,
you are well on your way to mastering relativity!

Before tackling simultaneity, let us first think about {\em
collocation}.  Two events (such as A and E in figure \ref{simul}) are
collocated if they have the same $x$ value.  However, collocation is a
concept that depends on reference frame.  For instance, George is
driving from Socorro to Albuquerque.  Just as he passes Escondida he
sneezes (event A in figure \ref{simul}).  As he drives by Belen, he
sneezes again (event D).  In the reference frame of the earth, these
two sneezes are not collocated, since they are separated by about
forty miles.  However, in the reference frame of George's car, they
occur in the same place --- assuming that George hasn't left the
driver's seat!

Notice that any two events separated by a timelike interval are
collocated in {\em some} reference frame.  The speed of the reference
frame is given by (\ref{slope2}), where the slope is simply the slope
of the world line connecting the two events.

Classically, if two events are simultaneous, we consider them to be
simultaneous in all reference frames.  For instance, if two clocks,
one in New York and one in Los Angeles, strike the hour at the same
time in the earth reference frame, then these events also appear to be
simultaneous to instruments in the space shuttle as it flies over the
United States.  However, if the space shuttle is moving from west to
east, i. e., from Los Angeles toward New York, careful measurements
will show that the clock in New York strikes the hour {\em before} the
clock in Los Angeles!

Just as collocation depends on one's reference frame, this result
shows that simultaneity also depends on reference frame.  Figure
\ref{simul} shows how this works.  In figure \ref{simul} events A and
B are simultaneous in the rest or unprimed reference frame.  However,
in the primed reference frame, events A and C are simultaneous, and
event B occurs at an earlier time.  If A and B correspond to the
clocks striking in Los Angeles and New York respectively, then it is
clear that B must occur at an earlier time in the primed frame if
indeed A and C are simultaneous in that frame.

The tilted line passing through events A and C in figure \ref{simul}
is called the {\em line of simultaneity} for the primed reference
frame.  Its slope is related to the speed, $U$, of the reference frame
by
\begin{equation}
     slope = U/c .
\end{equation}
Notice that this is the inverse of the slope of the world line
attached to the primed reference frame.  There is thus a symmetry
between the world line and the line of simultaneity of a moving
reference frame --- as the reference frame moves faster to the right,
these two lines close like the blades of a pair of scissors on the
$45^{\circ}$ line.

\subsection{Geometry of Spacetime}

\begin{figure}
\begin{center}
\psfig{figure=geom.eps,height=3in,width=3in,angle=0}
\end{center}
\caption{Resolution of a spacetime interval into different reference
frames.}
\label{geom}
\end{figure}

Figure \ref{geom} shows an interval $I$ between the events O and A in
spacetime resolved into space and time components in two different
ways.  In the unprimed reference frame the space and time components
are $X$ and $cT$, while in the primed frame they are $X'$ and $cT'$.
The spacetime interval is a quantity that doesn't depend on reference
frame, just as the distance between objects in ordinary space (i. e.,
the number of miles between Albuquerque and Socorro) doesn't depend
on the coordinate system used.  Thus, from (\ref{pytheqn}), we have
\begin{equation}
     I^2 = X^2 - c^2 T^2 = {X'}^2 - c^2 {T'}^2 .
\end{equation}

\begin{figure}
\begin{center}
\psfig{figure=dilation.eps,height=2.3in,width=5in,angle=0}
\end{center}
\caption{Two views of the relationship between three events, A, B,
and C.  The left panel shows the view from the unprimed reference
frame, in which A and C are collocated, while the right panel shows
the view from the primed frame, in which A and B are collocated.}
\label{dilation}
\end{figure}

The facts that 1) $I$ is independent of coordinate system and 2) $I$
has special meaning in certain cases can be used to attach a meaning
to $I$ in all circumstances.  For example, figure \ref{dilation} shows
two views of three events.  In the left panel the triangle A-B-C is a
right triangle oriented with the coordinate axes, with the interval
$I$ being the hypotenuse.  The Pythagorean theorem of spacetime thus
applies, and
\begin{equation}
     I = ( X^2 - c^2 T^2 )^{1/2} ,
\label{timeint}
\end{equation}
where $X$ and $T$ are defined in the figure.  The same triangle is
shown in the right panel from the point of view of a reference frame
that is moving to the right relative to the unprimed frame so as to
make events A and B collocated.  Notice that in this frame, the
triangle A-B-C is no longer a right triangle.  this is a consequence
of the geometry of spacetime --- angles are not preserved in
transformations between reference frames in this geometry, and
therefore have no particular physical meaning.  However, since A and
B are collocated in the primed frame, the interval between them is
simply $ic$ times the time between these two events, $T'$.

Rearranging (\ref{timeint}) and using (\ref{slope2}), which indicates
that the velocity difference between the reference frames is
$V = X/T$, we find
\begin{equation}
     icT' = I = (X^2 - c^2 T^2 )^{1/2} = T(V^2 - c^2)^{1/2} ,
\end{equation}
which reduces to
\begin{equation}
     T' = T [1 - (V/c)^2 ]^{1/2} .
\end{equation}
This is the famous {\em time dilation} effect of special relativity ---
a moving clock appears to run more slowly than normal to a stationary
observer.

\begin{figure}
\begin{center}
\psfig{figure=lorentz.eps,height=2.3in,width=5in,angle=0}
\end{center}
\caption{Definition sketch for understanding the Lorentz contraction.
The parallel lines represent the world lines of the front and the rear
of a moving object.  The left panel shows a reference frame moving
with the object, while the right panel shows a stationary reference
frame.}
\label{lorentz}
\end{figure}

A similar argument can be made to show how the postulates of
relativity result in the {\em Lorentz contraction}.  Figure
\ref{lorentz} compares the length of a moving object measured in its
own reference frame (left panel) with its length as measured in a
stationary reference frame (right panel).  The length of a moving
object is measured by {\em simultaneously} measuring the positions
of the front and the rear of the object and subtracting these two
numbers.  Events A and C correspond to these position measurements
for the stationary reference frame since they are respectively on
the rear and front world lines of the object.  Thus, the interval
AC, which is equal to $X$, is the length of the object as measured
in the stationary frame.

In the left panel, $X$ is the hypotenuse of a right triangle.
Therefore, by the Pythagorean theorem of spacetime, we have
\begin{equation}
     X = ({X'}^2 - c^2 {T'}^2 )^{1/2} .
\label{lorpthag}
\end{equation}
Now, the line passing through A and C in the left panel is the line of
simultaneity of the stationary reference frame.  The slope of this
line is $-V/c$, where $V$ is the speed of the object relative to the
stationary reference frame.  Geometrically in figure \ref{lorentz},
the slope of this line is $- cT' / X'$, so we find by equating these
two expressions for the slope that
\begin{equation}
     T' = V X' / c^2 .
\label{simslope}
\end{equation}
Finally, combining equations (\ref{lorpthag}) and (\ref{simslope})
results in
\begin{equation}
     X = X' (1 - V^2 / c^2 )^{1/2} .
\end{equation}
This says that the length of a moving object as measured in a
stationary reference frame ($X$) is {\em less} than the actual length
of the object as measured in its own reference frame ($X'$).  This
apparent reduction in length is called the Lorentz contraction.

\end{document}
