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\begin{document}
\title{Physics 51A: Modern Physics}
\author{Prof. Clare Yu\\
email: cyu@uci.edu\\
phone: 949-824-6216\\Office: 210E RH}
\date{Winter 2013}
\maketitle
\setcounter{page}{1}
\thispagestyle{empty}

\centerline{\bf CHAPTER 1} 
\centerline{Introduction}

Have you ever watched a horror or science fiction movie where everything
starts out very normal and ordinary, but then strange and bizarre things  
begin to happen? Well, special relativity is like that. Let me give you 
a brief preview. 

We live in a 3 dimensional world: length, width, height. Actually, it's 4
dimensional if you include time. For example, if you have a meeting to go to,
you need to know where and when it will be, i.e., you need to specify
the position (location) and the time. We tend to think of spatial dimensions
as very different from time. Positions are fixed in time and space
but time passes.
However, as we shall see, Einstein's theory of special relativity tells us 
that space can transform into time and time can be transformed into space. 
Weird, huh? Space and time are not as ordinary as they appear to be.
This turns out to be a consequence of the fact that the speed of light
is the universal speed limit; nothing (technically speaking, no information) 
can travel faster than the speed of light.

\centerline{\bf Static Reference Frames}

But let's start out with the boring and ordinary. (Pay attention
because things will slowly start to get weird and you won't even
realize it at first.) In particular, let's talk
about frames of reference. If we specify a position ${\vec r}=(x,y,z)$, we
are measuring $(x,y,z)$ with respect to a given frame of reference. That 
same position could have different coordinates in another reference frame.
For example, consider the point P in Figure 1. In one frame of reference
P has the coordinates $(x=3,y=4)$ while in the other (primed) reference frame,
P has the coordinates $(x^{\prime}=1,y^{\prime}=1)$. The position of
P is measured {\it relative} to a coordinate system.

\begin{figure}
\begin{center}
\leavevmode
\epsfxsize=2.0 true in
\epsfbox{TwoRefFramesStatic.eps}
\end{center}
\caption{Two Static Reference Frames}
\end{figure}

Similarly, when we measure the time of an event, it is with respect to 
when $t=0$. For example, when we say that the year is 2013, 
we are measuring with respect to the year 0 A.D. Of course, the
laws of physics hold and are the same (invariant) no matter where the origin of
the coordinates or the time $t=0$ is set. 

\centerline{\bf Moving Reference Frames}

Now let's consider 2 reference frames moving relative to one another 
with a constant velocity ${\vec v}$. Each reference frame has an observer.
For example, suppose reference frame S is fixed to the ground and
reference frame S' is fixed to a train moving along the tracks with
constant velocity ${\vec v}$. Each observer is standing still but thinks
the other observer is moving. The person on the train tracks sees
the train and the passenger moving at velocity ${\vec v}$. The
passenger on the train sees the person standing on the tracks go by
with velocity $-{\vec v}$. 

If the passenger is standing and drops a ball straight down, then
the ball will fall at his feet. The observer on the tracks sees the
ball follow a parabola and land to the right of where it was released.
This is because the ball is traveling horizontally with the same speed
as the train. (See Figure 1.2 in the book.)

If the passenger throws the ball at a velocity ${\vec u}^{\prime}$ 
relative to the train, then the observer standing on the ground sees
the ball traveling at a velocity ${\vec u}$ given by the sum
of the train velocity and the velocity of the ball with respect to the
train:
\be
{\vec u}={\vec u}^{\prime}+{\vec v}
\label{eq:addspeeds}
\ee
The book calls this equation the classical velocity-addition formula.
As we shall see, this doesn't hold in Einstein's Theory of Special Relativity.

The next question is ``Do Newton's Laws of Motion hold in moving
reference frames?''
As your book describes in Section 1.3, Newton's 3 Laws of Motion
hold in all reference frames that are moving at constant velocity
(including $v=0$) with respect to one another, i.e., in unaccelerated
reference frames. These reference frames are often called
{\bf inertial frames}. Let's review what these laws of motion are:
\begin{enumerate}
\item {\bf The law of inertia:} A body on which no external forces act 
moves with constant velocity.
\item ${\vec F}=m{\vec a}$
\item For every action, there is an equal and opposite reaction.
\end{enumerate}
So Newton's laws of motion are invariant in all inertial frames of reference.

\centerline{\bf The speed of light is $c$ in all inertial frames.}

What about the laws of electromagnetism, i.e., Maxwell's equations? Are they
invariant in inertial frames of reference? Classically, the answer is no,
but Einstein showed that the right answer is yes. To see this, let's 
consider the speed of light $c$.
\be
c=\frac{1}{\sqrt{\varepsilon_0\mu_{0}}}=3.00\times 10^8 \;{\rm m/s}
\label{eq:c}
\ee
where $\varepsilon_0$ and $\mu_0$ are the permittivity and the permeability
of the vacuum. These constants appear in Maxwell's equations. So if
Maxwell's equations hold in a frame $S$, then light travels the same speed $c$
in all directions. 

Now suppose a pulse of light is moving along the $+x$ axis in frame
$S$ at speed $c$. Consider another frame of reference $S^{\prime}$
moving along the $x$ axis with velocity ${\vec v}$. According to 
Eq. (\ref{eq:addspeeds}), an observer in $S^{\prime}$ would measure the
speed of the light pulse to be $c-v$. Similarly, if a pulse of light
were traveling at speed $c$ in $S$ in the $-x$ direction, then the
observer in $S^{\prime}$ would measure the speed of the light coming
toward him to be $c+v$. For light traveling at some oblique angle
with respect to the $x-$axis, the observer in $S^{\prime}$ would measure
the speed of light to be between $c-v$ and $c+v$.
Since Maxwell's equations demand that the speed
of light be exactly $c$ as given by Eq. (\ref{eq:c}), we conclude that
the laws of electromagnetism, unlike those of mechanics, are not valid
in frame $S^{\prime}$. 

This is what people believed before Einstein came along. They believed that
light propagated through a medium called ether which filled the vacuum, and
that there was just one frame, the {\bf ether frame}, in which light 
traveled at speed $c$. In any frame moving with respect to the ether, like
earth, light would be measured as traveling at some other velocity.

The obvious thing to do is to measure the speed of light experimentally. But
that's hard to do because light goes so fast. If you turn on a flashlight
and then ask someone with a stopwatch to see how long it takes for the light
to travel across the room, that would be impossible since the light travels
so fast. However, between 1880 and 1887, Albert Michelson, later
assisted by Edward Morley, measured the speed of light. Some of those
experiments were done near here, and that may be why we have a road called
Michelson. They figured that the speed of the earth with respect to the
ether was probably comparable to the earth's speed as it circled the sun,
i.e.,
\be
v\sim 3\times 10^4\;{\rm m/s}
\ee
This is about 1/10,000 the speed of light. So they expected to see the
speed of light differ from its value in the vacuum by about 1 part in $10^4$.
To do such a sensitive measurement, they built an interferometer that
split a beam of light into 2 beams that traveled perpendicular to each other,
and then later were deflected to intersect at one point. If the beams
traveled at different velocities or different distances, this could be
detected by constructive or destructive interference of the light waves.
Surprisingly, they found that light traveled at the speed $c$ in all 
directions. Later and more precise experiments have found the same result.
Light travels at the same speed in all directions in all inertial reference
frames. (More details about the Michelson-Morley experiment can be found
in your book in section 1.5.) This means that the laws of electromagnetism
are the same (invariant) in all inertial frames of reference.

This means that the classical velocity-addition formula 
(Eq. (\ref{eq:addspeeds})) doesn't hold for light. In other words, if
an observer on the ground shines light along the $+x-$axis, the observer
in the moving train measures the speed of the light as $c$, irrespective of
the train's velocity. Now things are starting to get weird. Einstein
was the first to work out the consequences of this startling finding, and
the result was the Theory of Special Relativity.

\centerline{\bf Postulates of Relativity}
First let us expand our definition of an inertial frame to be any reference
frame (that is, system of coordinates $x$, $y$, $z$, and time $t$) where
all the laws of physics hold in their simplest form. (Your book adds the
stipulation ``simplest form'' to exclude things like introducing
a ``fictitious'' centrifugal force so that the laws of statics hold in
an rotating frame.)

Now here are the postulates of relativity:
\begin{enumerate}
\item {\it If $S$ is an inertial frame and if a second frame $S^{\prime}$ moves
with a constant velocity relative to $S$, then $S^{\prime}$ is also an
inertial frame.} This means that there is no absolute motion, and that
everything is relative, so to speak. Going back to the train and the 
observers, the train is moving relative to the observer standing
next to the train tracks, and the ground is moving relative to the 
observer on the train. No inertial frame is better than any other
frame. 
\item {\it In all inertial frames, light travels through the vacuum with
the same speed, $c=299,792,458$ m/s in any direction.} This is the conclusion
of the Michelson-Morley experiment.
\end{enumerate}

\centerline{\bf Measurement of Time}
It turns out that the time between 2 events can be different in different
inertial frames, so we need to be
very careful about how time is measured. Basically we can think of
a reference frame $S$ filled with a team of observers, each with a clock
and all the clocks are synchronized to the same time. They can synchronize
their clocks by sending a flash of light at a specified time and take
into account the time it takes for the light to travel between clocks. Any
event that happens in frame $S$ will be noted by a nearby observer and
she can tell the other observers when the event happened according to her
clock.

\centerline{\bf The Relativity of Time: Time Dilation}

It turns out that the time between 2 events is different when viewed from
two different frames that are moving at a constant velocity relative to 
one another. So let's consider 2 inertial frames of reference: one anchored
to the ground that we'll call $S$, and the other in a train moving at velocity
${\vec v}$ relative to the ground. We'll call the train frame $S^{\prime}$.
In the train, a flash of light leaves the floor, travels up to the ceiling 
where it hits a mirror, and is reflected back to the floor. When it returns
to the floor, a beep is heard by the passenger on the train. The height
of the railcar is $h$, so the time for the light to make a round trip up
to the ceiling and back is 
\be
\Delta t^{\prime}=\frac{2h}{c}
\label{eq:trainTime}
\ee
This is the time interval recorded by a passenger in the train, i.e., 
in the frame $S^{\prime}$. (See Figure 2.) The light flash leaves and returns
to the same place on the floor of the train.

\begin{figure}
\begin{center}
\leavevmode
\epsfxsize=3.5 true in
\epsfbox{TimeDilation.eps}
\end{center}
\caption{Time Dilation}
\end{figure}

In the ground frame let us denote the time interval by $\Delta t$. The
ground observer would say that the flash leaves at one place and returns
to another because the train moved in the interval. (See Figure 2.)
We can apply Pythagoras' theorem to calculate the distance $c\Delta t/2$
from the flash on the floor to the ceiling as observed by the person on
the ground:
\be
\left(\frac{c\Delta t}{2}\right)^{2}=h^{2}+\left(\frac{v\Delta t}{2}\right)^{2}
\ee
Solving for $\Delta t$ gives
\be
\Delta t=\frac{2h}{\sqrt{c^2-v^2}}=\frac{2h}{c}\frac{1}{\sqrt{1-\beta^{2}}}
\ee
where
\be
\beta =\frac{v}{c}\leq 1
\ee
We will see $\beta$ a lot, so it's good to be familiar with it. 
Another variable that is used a lot is $\gamma$ defined as
\be
\gamma=\frac{1}{\sqrt{1-\beta^2}}=\frac{1}{\sqrt{1-(v/c)^2}} \geq 1
\ee
So we can write
\be
\Delta t=\frac{2h}{c}\gamma
\ee
Using Eq. (\ref{eq:trainTime}), we obtain
\be
\Delta t=\gamma\Delta t^{\prime} \geq \Delta t^{\prime}
\label{eq:timedilation}
\ee
It turns out that this equation applies to the interval between any 2 events
that occur at the same place in the frame $S^{\prime}$.
Let's look at this equation. Since $\gamma \geq 1$, it says that the ground
observer measures a longer time interval between 2 events
than the train based observer. Basically time runs slower on the train.
Notice, as a check, that if $v=0$ (the train is standing still), then 
$\Delta t=\Delta t^{\prime}$ which is what we expect.

The faster the train goes, i.e., the bigger $v$ is, the slower time is on
the train. If the train goes at the speed of light, then
\be
\beta = \frac{v}{c}=1
\ee
and
\be
\gamma=\frac{1}{\sqrt{1-\beta^2}}=\frac{1}{0}=\infty
\ee
The time interval measured on the ground would be infinity which is
absurd.  What if $v>c$? Then $\beta =v/c > 1$, and $\gamma$ would be 
imaginary, which is unphysical (impossible). 
The speed of light $c$ is the speed limit of the universe.
Nothing can go faster than the speed of light. (Technically, no 
information can go faster than the speed of light.) We shall see
that nothing with nonzero mass can travel at the speed of light.
Anything with mass must travel at a speed less than the speed of light.
So $\beta < 1$ and $\gamma \geq 1$.

What if $v=0.99c$? Then $\beta=0.99$, and 
\begin{eqnarray}
\gamma &=& \frac{1}{\sqrt{1-(0.99)^2}}\approx 7 \nonumber \\
\Delta t &\approx & 7\Delta t^{\prime}
\end{eqnarray}

The frame of reference where 2 events (in our case the flash and the beep)
occur at the same place is special. The time measured between the 2 events
that occur in the same place is called the {\bf proper time}. It is 
designated by $\Delta t_0$ with a subscript zero to indicate the events
occurred at the same place in this reference frame. This is always
the shortest time measured between these 2 events in any reference frame.
The time measured in any frame between these two events is always greater
or equal to $\Delta t_0$:
\be
\Delta t =\gamma \Delta t_0\geq \Delta t_0
\ee
This effect is called {\bf time dilation}.

We have seen that time goes slower in moving frames of reference. How much
slower does it go on a jet plane traveling at 500 miles/hour? Say you 
fly across the country on a 5-hour flight. How much slower does time
go for you compared to a friend on the ground? Use the fact that 
$c=186,000$ miles/sec and that there are 3600 seconds in an hour to find
$\beta=500/(3600\times1.85\times 10^5)=7.47\times 10^{-7}$,
$\gamma=1 + (\beta^2/2)=1+2.79\times 10^{-13}$. So the difference is
$(2.79\times 10^{-13})\times (5\; {\rm hours})\times(3600$ seconds/hour)=
$5.02\times 10^{-9} \approx$  5 ns which is imperceptible.

\centerline{\bf Experimental Evidence for Time Dilation}

There is good experimental evidence confirming time dilation and the
conclusion that time goes slower in moving reference frames. For example,
in 1971, jet planes carried atomic clocks around the world, and then
the clocks were compared to atomic ground based clocks. The predicted
difference was $275\pm 21$ ns, and the measured discrepancy was
$273\pm 7$ ns which is in good agreement.

Another confirmation is provided everyday by unstable subatomic particles
which disintegrate. When they decay is random, but the {\bf half-life}
$t_{1/2}$ is well known. The half-life is the time it takes for half
of a large sample of particles to decay.
For example, the charged $\pi$ meson, or pion,
is formed by collisions between rapidly moving atomic nuclei in particle
accelerators. Its half-life is 
\be
t_{1/2}=1.8\times 10^{-8}\;{\rm s}
\ee 
At Fermilab near Chicago, pions are produced moving very close to the speed of
light:
\be
v=0.9999995c
\ee
and they must travel about 1 km to the experimental area. Now if we didn't
know about relativity, how long (in lab time) would it take for them to get
there? 
\be 
T({\rm lab\;frame})=\frac{L}{v}=\frac{10^3\;{\rm m}}{3\times 10^8\;{\rm m/s}}=
3.3\times 10^{-6}\;{\rm s}\approx 183t_{1/2}
\ee
which implies that there won't be any pions left by time they get 1 km.

But let's see what relativity says. Note that
\begin{eqnarray}
\beta &= &0.9999995\nonumber\\
\gamma &=& 1000
\end{eqnarray}
So
\begin{eqnarray}
t_{1/2}({\rm lab\;frame})&=&\gamma t_{1/2}(\pi\;{\rm rest\;frame})\nonumber\\
&=&1000\times(1.8\times 10^{-8}\;{\rm s})\nonumber\\
&=&1.8\times 10^{-5}\;{\rm s}
\end{eqnarray}
So
\be
T({\rm lab\;frame})\approx 0.2t_{1/2}({\rm lab\;frame})
\ee
So in the lab, the pion's half-life increases by a factor
of $\gamma$ because the pions are moving so fast. 
The actual time of flight in the lab is only about 1/5 the relevant
half-life. So plenty of pions make it to the experimental area.

\centerline{\bf Length Contraction}

We have seen that time depends on the reference frame. It turns out that
length or distance does too. We can still use distance = velocity $\times$ 
time since this is just the definition of velocity.

We again consider 2 reference frames: one ($S^{\prime}$)
on a train moving with constant velocity ${\vec v}$ and the 
other ($S$) fixed to the ground. We want to measure
the length of the train. An observer on the ground measures the time $t_1$
when the front of the train passes him, and the time $t_2$ that the back
of the train passes him. Then he calculates the length $\ell$ of the train
to be
\be
\ell=v\times (t_2 - t_1)=v\Delta t
\label{eq:Lground}
\ee
What do observers on the train measure to be the length $\ell^{\prime}$
of the train? Let's put one observer at the front of the train and one at 
the back of the train. When the observer at the front passes a tree
on the ground, he records the time $t_1^{\prime}$. When the observer
at the back of the train passes that same tree, he records the time
$t_2^{\prime}$. So the observers on the train calculate the length 
of the train to be
\be
\ell^{\prime}=v\times (t_2^{\prime} - t_1^{\prime})=v\Delta t^{\prime}
\label{eq:Ltrain}
\ee
$\Delta t^{\prime}$ and $\Delta t$ are different so $\ell^{\prime}$ and
$\ell$ must be different. We need to relate $\Delta t^{\prime}$ and 
$\Delta t$ using the time dilation formula. In this case the front
of the train opposite the tree and the back of the train opposite
the same tree occur at the same place on the ground. So the time on
the ground is the proper time. Thus, we have
\be
\Delta t^{\prime}=\gamma\Delta t
\ee
Plugging this into Eq. (\ref{eq:Ltrain}) yields
\begin{eqnarray}
\ell^{\prime}&=&v\Delta t^{\prime} \nonumber\\
&=&v\gamma\Delta t\nonumber\\
&=&\gamma\ell
\end{eqnarray}
or
\be
\ell=\frac{\ell^{\prime}}{\gamma}\leq \ell^{\prime}
\ee
So the observers on the train and on the ground measure a different
length for the train, unless the train is standing still. The observers
on the ground measure a shorter train than the passengers on the train do.
The faster the train, the shorter the train according to the people 
on the ground.
If the train were moving at the speed of light, $\gamma = \infty$, and
the people on the ground would say the length of the train is zero which is
absurd. Again, nothing with mass can travel at or faster than the speed of
light.

When we measure length or distance in the rest frame of the object, we
call that length the {\bf proper length} $\ell_0$. 
The length of the train in the rest frame of the train is 
$\ell_0=\ell^{\prime}$, and we can write
\be
\ell=\frac{\ell_0}{\gamma}\leq \ell_0
\label{eq:lengthContraction}
\ee
The maximum length of an object is in its rest frame. All other frames
of reference measure $\ell\leq \ell_0$. This is called {\bf length
contraction} or {\bf Lorentz contraction}. Basically, the faster an object
moves, the shorter it appears to be in the direction of motion. Weird, huh?

\centerline{\bf Evidence for Length Contraction}

Length contraction is well established experimentally. Let's go back to 
the pion experiment where we saw that in the lab, the pions' half-life
was increased by a factor of $\gamma$. Suppose, however, that we go to
the pions' rest frame where the pions are stationary and there is no
increase in their half-life. How do they reach the experimental area?
From the point of view of the pions, the pipe or tunnel is moving,
and the length of the pipe is contracted by a factor of $\gamma$. So
if the length of the pipe is $L=1$ km in the lab, the pions measure it to be
$L/\gamma=L/1000=1$ meter. Of course, the number of pions surviving the
journey is the same in both reference frames.

\centerline{\bf Lengths Perpendicular to the Direction of Motion}

It turns out that lengths perpendicular to the direction of motion are
not changed. (see page 25 in your book for more details)
Only lengths parallel to the direction of motion experience
Lorentz contraction.

\centerline{\bf Lorentz Transformation}
Let us now consider a more general problem. If we know the coordinates 
$x$, $y$, $z$, and the time $t$ of an event measured in a frame $S$, what
are the coordinates $x^{\prime}$, $y^{\prime}$, $z^{\prime}$, 
and $t^{\prime}$ of the same event in the frame $S^{\prime}$?

Let us start by setting up the two frames of reference: $S$ anchored to the
ground and $S^{\prime}$ anchored to a train moving with velocity 
${\vec v}$ relative to the ground. We choose the $x-$axis and the
$x^{\prime}-$axis to lie along the same line, parallel to ${\vec v}$.
We further choose the origins $O$ and $O^{\prime}$ to coincide at time
$t=t^{\prime}=0$. Your book calls this arrangement the {\bf standard
configuration}. 

{\bf Galilean Transformation}: Let's start by giving the classical
answer. Time is the same in both frames, so $t=t^{\prime}$. 
Perpendicular coordinates are the same, so $y=y^{\prime}$ and 
$z=z^{\prime}$. Finally, $x^{\prime}=x-vt$ or 
$x=x^{\prime}+vt=x^{\prime}+vt^{\prime}$.
This is called the Galilean transformation (after Galileo Galilei,
the great Italian astronomer):
\begin{eqnarray}
x &=& x^{\prime}+vt^{\prime}\nonumber\\
y &=& y^{\prime}\nonumber\\
z &=& z^{\prime}\nonumber\\
t &=& t^{\prime}
\label{eq:Galilean}
\end{eqnarray}
This is the classical answer and is fine at low velocities much
smaller than the speed of light. But it is not correct relativistically
when speeds approach the speed of light. For example, it does not give
time dilation.

{\bf Lorentz Transformation}:
The correct relativistic transformation is called the Lorentz transformation.
We can derive it as follows. Suppose a pulse of light is emitted at
$t=t^{\prime}=0$ from the origin $O$ and $O^{\prime}$. (Recall that the
origins coincide at $t=t^{\prime}=0$.) In frame $S$ the equation of the
spherical wave front emitted at $t=0$ is
\be
x^2+y^2+z^2=c^2t^2
\label{eq:lightS}
\ee
In the frame $S^{\prime}$, the equation of the spherical wave front of 
light must be
\be
x^{\prime 2}+y^{\prime 2}+z^{\prime 2}=c^2t^{\prime 2}
\label{eq:lightSprime}
\ee
Note that the speed of light $c$ is the same in the two reference frames
and so is the same in Eqs.~(\ref{eq:lightS}) and (\ref{eq:lightSprime}).
If we substitute the Galilean transformation (Eq. (\ref{eq:Galilean}))
into Eq. (\ref{eq:lightS}), we do not obtain Eq. (\ref{eq:lightSprime}).
So let us try a solution of the form
\begin{eqnarray}
x{\prime} &=& \alpha x + \varepsilon t \nonumber\\
y^{\prime} &=& y\nonumber\\
z^{\prime} &=& z\nonumber\\
t^{\prime} &=& \eta t +\delta x 
\label{eq:Lorentztrial}
\end{eqnarray}
where we have to find the parameters $\alpha$, $\varepsilon$,
$\delta$, and $\eta$ that will make Eqs. (\ref{eq:lightS}) and
(\ref{eq:lightSprime}) consistent. To find these parameters,
we can use time dilation and Lorentz contraction to find that
$\alpha=\eta=\gamma$. Then plugging Eqs. (\ref{eq:Lorentztrial})
into Eq. (\ref{eq:lightSprime}) and demanding that we get
Eq. (\ref{eq:lightS}), we obtain the Lorentz transformation:
\begin{eqnarray}
x^{\prime} &=&\gamma(x-vt)\nonumber\\
y^{\prime} &=&y\nonumber\\
z^{\prime} &=&z\nonumber\\
t^{\prime} &=&\gamma\left(t-\frac{vx}{c^2}\right)
\end{eqnarray}
The inverse Lorentz transformation is
\begin{eqnarray}
x &=&\gamma(x^{\prime}+vt^{\prime})\nonumber\\
y &=&y^{\prime}\nonumber\\
z &=&z^{\prime}\nonumber\\
t &=&\gamma\left(t^{\prime}+\frac{vx^{\prime}}{c^2}\right)
\end{eqnarray}

This can be viewed as 4 equations and 4 unknowns.
In many problems, you are given 3 unknowns and you have to find the
fourth.

Let's take a moment to look at these equations. Space ($x$) in one
frame is a combination of space ($x^{\prime}$) and time ($t^{\prime}$)
in another inertial frame.
Time $t$ in one frame is a combination of space ($x^{\prime}$) and 
time ($t^{\prime}$) in another inertial frame. We think of space and
time as separate things but they are actually mixed together. Space-time
is the fabric of the universe.

\centerline{\bf Example of the Lorentz Transformation: Length Contraction}
Let's start with the Lorentz transformation and derive the Lorentz
contraction. The length $\ell_0$ of a moving train according to people
on the train is
\be
\ell_0=\ell^{\prime}=x^{\prime}_2-x^{\prime}_1
\ee
Let's plug in the Lorentz transformation:
\begin{eqnarray}
\ell_0&=&\ell^{\prime}=x^{\prime}_2-x^{\prime}_1\nonumber\\
&=&\gamma\left(x_2-vt_2\right)-\gamma\left(x_1-vt_1\right)\nonumber\\
&=&\gamma\left(x_2-x_1\right)-\gamma v\left(t_2-t_1\right)\nonumber\\
&=&\gamma\Delta x-\gamma v\Delta t
\end{eqnarray}
On the ground, two observers measure the coordinates of the front and back
of the train at the same time $t$. So $\Delta t=0$.
They measure the length of the train as
\be
\ell = x_2-x_1
\ee
So we can write
\be
\ell_0=\gamma\ell
\ee
We recover the length contraction equation
Eq. (\ref{eq:lengthContraction}).

\centerline{\bf The Velocity-Addition Formula}
The classical velocity-addition formula relates the velocity ${\vec u}$
of an object in frame $S$ to the velocity ${\vec u}^{\prime}$ in $S^{\prime}$:
\be
{\vec u}={\vec u}^{\prime}+{\vec v}
\ee
This is Eq. (\ref{eq:addspeeds}). This equation is not valid relativistically.
The speed of light in the moving train (or moving frame of reference)
is the same as in $S$, the ground based frame, but this equation predicts
that light should have different velocities in different frames. Michelson
and Morley showed that light travels at the same velocity in all inertial
frames. In section 1.13, your book starts from the Lorentz tranformation
and uses the basic definitions of velocity
\be
u_x=\frac{\Delta x}{\Delta t},\;\quad
u_y=\frac{\Delta y}{\Delta t},\;\quad
u_z=\frac{\Delta z}{\Delta t}
\ee
to show that the correct relativistic velocity addition formulas are
\begin{eqnarray}
u^{\prime}_x &=&\frac{u_x-v}{1-u_xv/c^2}\nonumber\\
u^{\prime}_y &=&\frac{u_y}{\gamma\left(1-u_xv/c^2\right)}\nonumber\\
u^{\prime}_z &=&\frac{u_z}{\gamma\left(1-u_xv/c^2\right)}
\end{eqnarray}
Notice that in the nonrelativistic limit when $v\ll c$, $\gamma\approx 1$,
and we recover the classical velocity addition formula:
\begin{eqnarray}
u_x^{\prime}&\approx& u_x-v\nonumber\\
u_y^{\prime}&\approx& u_y\nonumber\\
u_z^{\prime}&\approx& u_z
\end{eqnarray}

\end{document}

Since the
proper time and proper length were in $S^{\prime}$, it is easier
to use the inverse transformation going from $S^{\prime}$ to $S$:
\begin{eqnarray}
x &=& \alpha x^{\prime} - \varepsilon t^{\prime} \label{eq:Lorentzxprime}\\
y &=& y^{\prime}\nonumber\\
z &=& z^{\prime}\nonumber\\
t &=& eta t^{\prime} -\delta x^{\prime} \label{eq:Lorentztprime}
\end{eqnarray}
So to use time dilation, think of the flash and beep taking place on the
floor of the train, i.e., at the same place ($\Delta x^{\prime}=0$):
\begin{eqnarray}
\Delta t&=&t_2-t_1=\eta\left(t_2^{\prime}-t^{\prime}_1\right)-
\delta\left(x_2^{\prime}-x_1^{\prime}\right)\nonumber\\
&=&\eta\Delta t^{\prime}-\delta\Delta x^{\prime}\nonumber\\
&=&\eta\Delta t^{\prime}
\label{eq:timeDilLT}
\end{eqnarray}
Comparing this to Eq. (\ref{eq:timedilation}), we see that $\eta=1/\gamma$.

For an event taking place at the same time ($\Delta t^{\prime}=0$), 
e.g., like measuring the length of the train on the train, i.e., in the
frame $S^{\prime}$:
\begin{eqnarray}
\Delta x &=&x_2-x_1=\alpha\Delta x^{\prime}-\varepsilon\Delta t^{\prime}
\nonumber\\
\Delta x &=&\alpha\Delta x^{\prime}
\end{eqnarray}

