11\. Relativity

The nature of time

What is time? Time moves. But what does *that* mean? Does time ever slow
down? And if it did, would we notice? Could we notice? What is it that
sets the \"pace\" of time? Nothing about time is obvious. Is time a
universal concept, that exists throughout the Universe? Can we go back
in time? Can time run backwards? What determines the direction of time?

I can\'t answer all of those questions. Yet, given the mysterious nature
of time, you may be surprised at some of the things we do know about it.
Here is the strangest of them all:

If two twins are exactly the same age, and one travels while the other
stays at home, then when they are brought back together \-- the traveler
will have experienced less time than the other.

There is nothing odder about time than that. Yet Einstein gave us a
formula that tells us precisely how much less time the moving twin
experienced. And that fact has been experimentally verified with very
sensitive clocks flown on airplanes. Even radioactive atoms, when they
move, experience less time than those that are stationary. That fact is
verified every day at accelerator laboratories where such atoms are sent
near the speed of light, and physicists note that their radioactive
decays slow down.

The nature of time -- and of space -- is at the heart of the Theory of
Relativity. That\'s what this chapter is about. The Theory of Relativity
was created by Einstein in the early 1900s. It consisted of two parts.
The first part was called \"the Special Theory of Relativity\" and it
had to do with the nature of time, space, energy, and momentum. It was
in this work, published in 1905, that Einstein came up with the famous
equation E = mc2. The second part was published in 1916 and was called
\"the General Theory of Relativity.\" It was really a theory of gravity.
It \"explained\" all of gravity as due to a bending of space and time.
It is the theory that is needed to understand some of the recent
discoveries in cosmology about the nature of the Universe. Most of this
chapter will be about Special Relativity.

This chapter departs a bit from my previous philosophy. Future
presidents don\'t really need to know about the Theory of Relativity. It
is important, however, to physicists, to philosophers, to those who plan
trips to other planets, and to anyone who wants to have their mind
stretched beyond what this course has already done.

The fourth dimension

We return to the question, \"what is time?\" The trivial answer that you
will hear from some people is that time is \"the fourth dimension.\"
This sententious statement means less than it seems.

In this answer, the word \"dimension\" is being used in a very technical
and narrow way. If you want to specify a *location*, you have to give
three dimensions. They can be latitude, longitude, and altitude, or they
can be x, y, and z. But given a coordinate system, three numbers are all
you need to say precisely where an object is in that coordinate system.

If you want to specify an *event*, then it is sufficient to give the
location and the time of the event. Suppose I were to tell you that
there is an event at my house at 8 pm tonight. Then there is no
confusion; you might not know what is going to happen, but you have
located it in both time and space. The event can have a name, such as
\"Rosemary\'s birthday party\" or \"Muller goes to bed.\" But to be
unique (Rosemary has a birthday each year, and Muller goes to bed almost
every night) you also specify the time. This is what makes time the
fourth dimension, although in a trivial way. It is only the forth
dimension because you are specifying an event rather than just a
location.

That is not what is interesting about time. What is interesting is that
the amount of time can change depending on the velocity that an object
is moving. That idea requires some explanation.

Time dilation

I described in the opening of this chapter how two twins can experience
different amounts of time. That seems to violate common sense. How can
it be true? The answer is that the effect is very small unless the
velocity is very fast \-- that\'s why you never noticed it. In fact, it
is difficult even to measure the effect unless the velocity is near the
speed of light. For airplanes moving at 675 miles per hour, the effect
is about 5 x10^--13^. That means that in one day you would lose 44
nanoseconds. (That is the time it would take light to travel 44 feet.)
If you fly for a *year*, you will experience 16 microseconds less time
than your twin who doesn\'t travel.

This \"small\" effect becomes large if the velocity approaches the speed
of light. Suppose the second twin (Mary) moves extremely fast: at 60% of
the speed of light. How much will her time slow down? Einstein gave an
exact formula for calculating this. The answer turns out to be that
Mary\'s time will go slower by a factor of exactly 1.25. So if she
experiences a year, her stay-at-home twin (call him John) will
experience 1.25 years. She will be one year older, and he will be 1.25
years older. When they compare ages, John will be a quarter year older
than Mary. Yet they are twins, born at the same time.

No astronaut has ever traveled this fast, so it has never been observed
for people. But we have sent radioactive atoms that fast, and their
radioactivity does slow down, by exactly the Einstein factor.

The Twin Equation

The equation that says exactly how much time a moving person (or a
moving radioactive atom) experiences is remarkably simple. The amount of
time experienced by a moving person (or clock or radioactive atom), is
less by the time factor[^1]

ƒ = sqrt\[1-(v/c)^2^\]

The \"sqrt\" in this equation stands for \"square-root\", and the letter
c stands for the velocity of light in a vacuum. This is the equation I
used for the previous examples. If the velocity is 60% the speed of
light, then v/c = 0.6. Plug that in, and you\'ll get ƒ = sqrt\[1-0.36\]
= sqrt\[0.64\] = 0.8 = 80%. Try evaluating the function yourself. Here
is a list of the result you get for different values of v/c:

  --------- ---- ------ ------ ----- ----- ------ ------- ------ ------- --------- ---- --
  **v/c**   0    0.25   0.5    3/5   4/5   0.9    12/13   0.99   0.999   0.99999   1    

  ƒ         1    0.97   0.87   4/5   3/5   0.44   5/13    0.14   0.045   0.0.004   0    
  --------- ---- ------ ------ ----- ----- ------ ------- ------ ------- --------- ---- --

>  

The time factor ƒ depends only on the velocity. Once you know it, then
you have the twin equation. Let T~M~ be the time experienced by the
moving twin, and T~S~ be the time experienced by the stationary twin.
Then:

T~M~= ƒ T~S~

Note to the student about math. This chapter has a lot of math in it.
However, you will not be required to reproduce this math. The important
thing are the conclusions. The math is there largely for those who love
math, and want to see how the various quantities are all linked.

**Optional: for those who like to play with math:** Maybe you noticed
that some of the values in the table are given as exact fractions,
rather than in terms of their decimal approximations. You can verify
that if you have an exact right triangle, e.g. a 3:4:5 triangle, or a
5:12:13 triangle, then values of velocity given by fractions made of
these numbers will give exact fractions for the time factor ƒ function.

How can time depend on velocity?

How can time depend on velocity? To do that \-- sounds absurd. It goes
against intuition. It goes against everything we experience.

Or does it? Why do you believe that time is independent of your path?
Did you always believe it? I\'ll bet that you didn\'t believe it when
you were a child. An hour the dentist\'s office didn\'t seem to go as
fast as an hour in a swimming pool. But you were trained to watch
clocks, and to \"be on time\" and you finally learned that there is a
\"universal\" time that you can follow in order to get to appointments
on time. But it was never intuitively obvious.

Nor is it true. It is *almost* true, however, since for everyday
velocities, the gamma function is very close to having a value of 1, the
value for which there is no time dilation. Even in the airplane example,
the factor ƒ was very close to the value of 1. At 675 mph, v/c is
10^-6^. The factor (v/c)^2^ is 10^-12^. The time factor ƒ, on most
calculators, will come out to be exactly 1 \-- since they don\'t handle
enough decimal places! If done on a sufficiently accurate computer the
factor ƒ = 0.9999999999995. (There should be 12 nines in that number, if
I entered it correctly.) It can be written as 1 -- 5 x10^-13^.

That is pretty close to 1, so it is hard to notice the difference.[^2]
Nevertheless, in 1972, two scientists recognized that clocks had become
accurate enough that the twin effect could be measured even in an
airplane, Their results were published in Science Magazine.[^3] Their
results confirmed that even at the \"slow\" velocity of airplanes, the
equation works. The \"moving\" clock experienced less time than the
stationary clock on the ground.[^4]

Not all motion is relative

Isn\'t all motion relative? Who is to say which clock is moving? I raise
this issue only because it is a favorite complaint made by people who
have studied a little bit of relativity theory. In fact, in relativity
theory, it is *not* true that all motion is relative. The clock that is
moving is the one that had to be accelerated to make it return home.
[^5]

Optional: Here are some details, for those interested. The twin equation
is good only if applied in a single \"inertial\" frame of reference,
i.e. one in which there is no acceleration without applied forces. The
frame of the stay-at-home twin can be used. The frame of the traveling
twin cannot be used, because that twin must change direction half-way
through the trip. Which twin changed velocity? That is unambiguous,
since it takes force to accelerate.

Notation: beta and gamma

If you read other books on relativity, you will probably run across the
symbols β and γ. It is worthwhile to point out that these symbols, when
used in relativity theory, have nothing to do with the beta rays or
gamma rays. β is the Greek letter beta, and when used in relativity
theory it usually represents v/c:

β = v/c

Beta looks like a B with a tail. At everyday velocities it is close to
zero. At very high velocities, it can approach 1. The function ƒ can be
written in terms of beta:

ƒ = sqrt(1-β^2^)

γ is the Greek letter gamma. In relativity theory, it usually stands for
1/ƒ:

γ = 1/ƒ = 1/sqrt(1-β^2^)

 

When written on the blackboard, γ looks like a fish that is swimming
straight down.

If you are willing now to believe that time depends on velocity, then
you are ready to move on to a few more of the astonishing facts of
relativity theory. You need no more math. All the results use the same
function ƒ.

Length contraction

When an object moves to a velocity v, it gets shorter. Its new length is
the old length multiplied by the factor ƒ. This effect is called the
\"Lorentz contraction.\" It is named after the person who first proposed
that is was true, even before Einstein. (Einstein published his theory
of relativity in 1905. But it was based on two decades of work preceding
it, including that of Lorentz.) Just to make it explicit, if the
stationary object has rest length L~S,~. When the object moves, its new
length is L~M~ given by:

L~M~ = ƒ L~S~

The moving object is shorter, since its length is the rest length
multiplied by ƒ.

This contraction turns out to be tricky to measure. If your ruler is
moving with you, then it shortens too, so you think your object hasn\'t
gotten any shorter! To see the effect, you have to have a stationary
ruler measuring a moving object.

Remember, the factor ƒ is so close to 1 at everyday speeds, that this
effect is difficult to notice. It becomes really important only when v
approaches c, which occurs for particles emitted from radioactive
nuclei, in accelerators (popularly called \"atom smashers\"), and in
cosmology where distant galaxies are moving away from us at speeds that
approach the speed of light.

Velocities don\'t just add

Suppose you are moving at half the speed of light, c/2, and crash into
someone coming from the opposite direction who is also moving at half
the speed of light. What is your relative velocity? You would probably
expect it to be c, the sum of the two velocities.

But Einstein recognized that his assumption ignores the strange behavior
of time and lengths. After all, velocity is distance divided by time.
When he compensated for both of these phenomena, he derived the relative
velocity formula:

V = (v1 + v2)/(1 + (v1 v2/c^2^)

You are not required to know this formula, but I wanted to write it down
so I can show you the consequences. Watch what happens when you put v1 =
0.5 c and v2 = 0.5 c. You get

V = (c/2 + c/2)/(1 + c/2 c/2 /c^2^)

= c/(1 + 1/4)

= 0.8 c

So the relative velocity is less than c, i.e. it is less than the speed
of light. Suppose we try v1 = 0.9 c and v2 = 0.9 c? The answer is that
the relative velocity is still less than c! Try it yourself. The answer
I got is V = 0.994 c. In fact, no matter how close the individual
velocities are to c, their relative velocity is still less than c. (If
you like math, you might enjoy trying to prove that.)

One consequence of this is that you cannot get velocities equal to or
greater than c by adding together smaller velocities. For example, if
you have a multistage rocket, with the first stage getting to 0.9 c, and
the second stage getting to 0.9 c relative to the first, the total
velocity that you will get for the second stage is only 0.994 c.

invariance of the speed of light

Suppose the particle that is coming towards you is a photon, traveling
at the speed of light c. You move with a velocity v towards that photon.
With what velocity will the photon hit you when you are moving? It is
natural to assume it will be greater than c, but that is not true. I\'ll
use the velocity equation, with my velocity v1 = v, and the photon
velocity v2 = c. (Skip the equations if they confuse you, and go right
to the last paragraph in this section.) This gives the new photon
velocity to be:

V = (v1 + v2)/(1 + (v1 v2/c^2^)

= (v + c)/(1 + v c/c^2^)

= (v + c)/(1 + v/c)

Now I multiply the numerator and denominator by c:

V = c(v+c)/c(1 + v/c)

= c(v+c)/(v+c)

= c

So the photon hits me with velocity c. Even though I am approaching it,
it still hits me with the same speed c. This surprising property is
sometimes called \"the invariance of the speed of light.\" In fact, in
many books it is taken as a fundamental assumption, and then all the
rest of relativity theory can be derived from it.

Energy and mass

Einstein noticed another consequence of the velocity equation. The old
concept of momentum conservation no longer worked. If two objects had
equal and opposite momentum (mass times velocity; see chapter 3), then
on a collision the result would be at rest. Momentum was conserved. But
when he calculated the momentum for objects in which both were moving,
that was no longer true. Einstein guessed (correctly, it turns out) that
momentum conservation was correct, and the velocity equation was
correct. The mistake was that the mass of an object is not really
constant, but that it depends on velocity. If we let m~0~ be the mass of
the object when it is at rest, then its mass when it is moving, its
\"kinetic mass,\" is

m = m~0­­­­~/ƒ

= γ m~0~

So kinetic mass is *bigger* than rest mass by the same factor ƒ that
tells us how much length and time are *smaller*. For mass, we divide by
ƒ rather than multiplying.

When something gets a bigger mass, that means that it gets harder and
harder to accelerate, and it also means that the pull of gravity will
increase. But it also has an important consequence for energy.

E = mc^2^

Einstein took these calculations one step farther and calculated the
energy of a moving object. He deduced that it is given by:

E = m~0~ c^2^/ƒ

= γ m~0~ c^2^

If I restate this by substituting the kinetic mass m for m~0~, then this
equation becomes the most famous equation in all of relativity theory:

E = m c^2^

At first look, it appears that the energy of an object does not depend
on velocity! But that isn\'t correct, because the mass depends on
velocity. (The mass m = m~0~/ƒ = γ m~0~.) Nevertheless, the equation
looks very different from the old kinetic energy equation:

E = (1/2) m~0~ v^2^

How can the two equations be reconciled? At first look, it seems
impossible. At zero velocity, the kinetic energy equation gives E = 0,
whereas the Einstein\'s equation gives E = m~0~c^2^. Those are very
different numbers. Because the value of c is huge, the zero velocity
energy m~0~c^2^ is very huge.

Even though they disagree, the two equations are more similar than you
might think. For small velocities, it is possible to show that if the
velocity is low, then the Einstein equation can be approximated as
follows:

E ≈ m~0~ c^2^ + (1/2) m~0~ v^2^

(If you are mathematically inclined, you might try to prove this. I give
some hints in a footnote.[^6])

This approximate version of Einstein\'s energy equation, valid only at
low velocities, says that the energy is equal to the old term (1/2
mv^2^) plus a new term that is constant: m~0~ c^2^. This constant term
has a famous name. It is called the *rest energy.* The smaller part, the
1/2 mv^2^ term, is still called the *kinetic energy*. We now say that
the total energy is the sum of the rest energy plus the kinetic energy.

This equation says that even a particle that is at rest stores enormous
energy, roughly the same energy it would have (by the previous equation)
if it were moving near the speed of light. But how do you extract this
energy? Part of it is extracted when we have a radioactive decay. The
mass of the debris is less than the mass of the original particle, since
some of the energy has been turned into energy. But usually it just
stays there, playing no role in energy conservation since it doesn\'t
change. That\'s why it was never noticed.

turning mass to energy -- and energy to mass

With the discovery of radioactivity, the presence of the relationship
between mass and energy has been measured. When a radioactive particle
decays, the sum of the masses of the pieces is less than the mass of the
original atom. The energy of the radioactive explosion came from
converting mass to kinetic energy.

How much rest energy is there in a particle? For a proton, the rest
energy is 938 MeV. That is huge compared to the typical 1 MeV energy
released in radioactive decay. But notice that the rest energy of the
electron is not huge: only 0.511 MeV. That is why electrons can be
*created* in beta decay. (Recall the analogy I made in Chapter 4 -- the
electron is not inside the nucleus, any more than a sound wave is inside
your body waiting for you to speak. Both are created at the moment they
\"come out\".)

Some particles have all of their energy in kinetic energy. An example is
the photon. All of the energy of a photon can be absorbed when that
photon hits an object.

Physicists usually say that the \"rest mass\" of a photon is zero. That
is an odd statement, since you can\'t bring a photon to rest. But if you
take energy away from a photon (perhaps by scattering it off an
electron) then the energy can be made to get smaller and smaller.
Eventually the photon has energy approaching zero, and that could only
be true if it had no rest energy, i.e. had zero rest mass.

antimatter engines

Could we release the rest energy of the electron, and turn it into
kinetic energy? Yes, there is one way: use antimatter. An anti-electron,
also called a *positron*, has the same mass as the electron but opposite
charge. Bring it together with an electron, and the charges will cancel,
and all of their mass energy will be released. Two photons will emerge
in equal and opposite directions, and all the energy of photons is
kinetic (since they have zero rest mass -- as we will see in the next
section). Therefore all that energy can be turned into heat.

You can do much better if you bring a proton together with an
antiproton. The process of releasing all this energy is called
annihilation. When a proton is annihilated, virtually all of its 938 MeV
rest energy is released. That\'s why antimatter drives are so popular in
science fiction stories. Matter and antimatter may constitute the
ultimate energy fuel.

Kinetic energy can also be turned into mass. When a gamma ray passes
close to a nucleus, we often observe a phenomenon called \"pair
production.\" The energy of the gamma ray is suddenly converted into the
mass of a particle and an antiparticle, such as an electron and
positron. This is a fairly common occurrence for high energy gamma rays,
and it is seen (for example) in cosmic radiation. The first positron
ever detected was one that had been produced by a cosmic gamma ray.
Other collisions (such as between electrons) can also create such pairs.
The first anti-proton ever detected was created as part of a
proton-antiproton pair in the Berkeley atom smasher known as the
\"Bevatron\" located at the Lawrence Berkeley Laboratory.

The photon has zero rest mass

Here is another way to deduce that a photon has zero rest mass. The
equation E = mc^2^ = γ m~0~c^2^ is true for all particles. Remember that
γ = 1/ƒ. When the velocity v = c, then ƒ = 0, and this makes γ infinite.
This seems to say that any particle that travels at the speed of light,
such as a photon, must have infinite energy! But no -- that result
isn\'t right. The equation contains the rest mass m~0~. The rest mass of
the photon must be zero. That is the only number which, when you
multiply it by infinity, gives a non-infinite answer. In fact, zero
times infinity is \"indeterminate.\" That means that it could be any
number.[^7] You can\'t tell what it is unless you have another equation.
For the photon, we do have such an equation: from Chapter 10 (Quantum
physics) we have E = h f, where h is Planck\'s constant, and f is the
photon frequency.

Likewise, we conclude that if a particle has mass = 0, but has energy,
then it must move at the velocity c. It can\'t move faster, and it
can\'t move slower. It\'s energy is not related to its velocity, but
only to its wave frequency.

But what about light in glass? Doesn\'t that move at c/n, where n is the
index of refraction. Isn\'t that slower than c?

The answer is that the equation for the energy did not include the
glass; if you put that term in, then you no longer conclude that light
must travel at c. But in a vacuum, a zero mass particle must move at c.
All the time.

You might notice that the photon kinetic mass m = γ m~0~ is not zero. It
is only the rest mass m~0~ that is zero. That is odd, given the fact
that the photon cannot be brought to rest! Nevertheless, the constant
m~0~ in the equation that defines the rest mass, E = γ m~0~c^2^, must be
zero for the photon.

The fact that the photon has non-zero kinetic mass suggests that it
feels the force of gravity. In fact, Einstein\'s theory of gravity
called \"General Relativity\" predicts that photons will be affected by
gravity.

Massless particles have no time

The time dilation factor for a massless particle is still ƒ, but for
such a particle ƒ is zero! That means that while an hour passes for you
and me, the time that passes for a massless particle is zero. (If this
bothers you, then lets just assume that the particle has a very tiny
mass. It is moving at *almost* the speed of light. Then ƒ is very close
to zero. So the particle experiences very little time.

If a massless particle does not experience time, can it undergo
radioactive decay? Think about this for a moment. What would you guess?
The answer is: no. Massless particles, which have no internal time,
cannot radioactively decay.[^8]

Do neutrinos have mass?

The neutrino is a particle that we always thought had mass zero. They
are emitted in many nuclear decays, and they travel through most matter
without being noticed. Neutrinos are emitted from the sun, but sensitive
experiments to detect them have shown that we detect only 1/3 of the
number we expect. What is going on?

New experiments have been done that suggest an answer: the neutrinos are
changing, from ordinary neutrinos, into exotic neutrinos called \"muon
neutrinos\" and \"tau neutrinos.\" (Ordinary neutrinos are often called
\"electron neutrinos, since they are never created by themselves, but
always in combination with electrons.)

But if the neutrinos are changing, they must be experiencing time. That
means that they aren\'t traveling at the speed of light. That means that
they aren\'t truly massless.

So the observing a low number of solar neutrinos, physicists were led to
conclude that neutrinos must have some mass. That was a conclusion that
surprised nearly everyone.

Why you can\'t go to light speed

Suppose you take a massive particle (such as an electron, or a person)
that has a non-zero rest mass m~0~, and you accelerate it to the speed
of light. Then its energy is

E = γ m~0~ c^2^ = ∞

For any particle that has non-zero rest mass (e.g. you), if you
accelerate it to the speed of light c, then its energy would be
infinite. That means that it would take infinite energy to reach c. That
is the fundamental reason why you can\'t travel at light speed.

E = mc^2^ and the atom bomb

Many people think that Einstein\'s discovery of the equation E = m c^2^
led to the invention of the atomic bomb. In fact, this equation is
irrelevant to the bomb. In the late 1800s it was discovered that
radioactive decay released a million times more energy than chemical
explosions. That discovery was the key to nuclear energy. All that was
really needed was the discovery of a suitable chain reaction, and that
wasn\'t found until the late 1930s. Einstein\'s equation (which he
published in 1905) showed that the enormous release of energy would be
accompanied by a slight disappearance of mass. But nobody needed to know
that in order to build an atomic bomb.

Beyond light speed: tachyons

I showed that you can\'t send particles at the speed of light, because
that would take infinite energy. But can you send them faster? The
surprising answer is: maybe. If such particles exist, we call them
*tachyons*. They have a surprising property: they must travel faster
than light! They have lowest energy when they move at infinite speed,
and infinite energy when they approach light speed c.

optional section: the math of tachyons

Look at the Einstein energy equation: E = mc^2^/ƒ. The factor ƒ is

ƒ = sqrt\[1 -- (v/c)^2^\]

and it goes to zero when v = c. But suppose v is greater than c. Then ƒ
is the square-root of a negative number. That makes the energy factor ƒ
imaginary, right? Right. Doesn\'t that make the energy E imaginary too?

The surprising answer is: not necessarily. Suppose there were a particle
that had imaginary mass.

I know -- now you are completely confused. Does a particle with
imaginary mass mean that the particle doesn\'t really exist? Is it
imaginary? If you\'ve never felt comfortable with \"imaginary\" numbers,
don\'t worry. Stick with me, and learn the conclusions, if not the logic
that we use to get there.

The two imaginary numbers, the mass and ƒ, when multiplied together,
create a real number. That means that everything works out OK. We can
re-write the energy equation in terms of real numbers. Let the mass of
the tachyon be (*i* m~0~), where *i* is the square-root of --1, and m~0~
is a real number. Then, for a tachyon, the two values of *i* will
cancel, and we\'ll get:

E = m~0~ c^2^/sqrt\[(v/c)^2^ -- 1\]

This is the tachyon energy equation. Note that when v becomes infinite,
the energy E goes to zero. And when v is equal to c, the energy E is
infinite.

If you think that is weird, be assured that everyone else thinks so too.
But that doesn\'t prevent it from being true. Of course, as of the
writing of this book, no tachyons have been discovered, so they may not
exist.

More about tachyons

As I said, a tachyon is a particle (not yet discovered, although people
have looked) that travels faster than the speed of light. But how can
you get it going that fast? Doesn\'t it have to start at rest, and so it
would take infinite energy to get just up to c -- and therefore it could
never go beyond? No -- tachyons are assumed to be *born* fast. Photons
are never accelerated up to light speed; they start out that way. If
tachyons can be created, when they are created they already have speed
faster than c. They can lose energy by speeding up, or gain energy by
slowing down. (Note that that is backwards from the way ordinary
particles such as electrons behave.)

Tachyons may exist, and from time to time, physicists set up experiments
to search for them. However, I do not think they will find them. The
reason has to do with another property of relativity, having to do with
the simultaneity of events.

Simultaneity

Different people can experience different amounts of time. That means
that there can be no such thing as a universal clock, that tells
everyone what time it is. Your time depends on your motion.

One consequence of this fact is that it is fundamentally impossible to
determine whether two events are simultaneous.[^9] And it gets even
worse. The order in which two events happened can depend on the motion
of the observer.[^10] This happens with tachyons. Suppose you emit a
tachyon at point A, and it travels to point B, where it is absorbed. To
a different observer, one moving along the line between the two events
with high speed, the order will be reversed. In that moving reference
frame, event B will occur before event A. [^11]

This fact may bother you more than anything else in this chapter. It
means that, to one observer, event A precedes B, but to another
observer, event B comes first. Suppose the tachyon was shot from a gun,
and used in a murder. The shooter is arrested. In his defense he points
out that, in a moving reference frame, the victim was killed before the
gun was fired.

There is nothing in that story that violates the laws of physics. But it
does violate our sense of free will. If the victim dies, can we still
choose not to fire the gun? Physics doesn\'t answer this question. But
this violation of causality is enough to cause many physicists to
believe that tachyons probably do not exist.

Likewise for time travel. Can you go \"backwards\" in time? If you could
travel faster than the speed of light, then according to the equations
of relativity, the order of events can reverse. Does that mean you can
travel back in time? Well -- the first problem is getting yourself to go
faster than the speed of light. It takes infinite energy just to get up
to c. But -- could you tunnel across, go right to hyper-light speed
without ever being at light speed, by doing some sort of tunneling?
Physics can\'t rule that out. But the problem, as with the tachyon gun,
may be the havoc such events would play on our own concept of causality
and free will. That is enough to get some physicists to guess that
travel backwards in time will turn out to be impossible.

Pondering time

Having gotten this far in the theory of relativity, you have earned the
right to speculate along with the best physicists. Just to get you
started, here are some Questions to Ponder:

Is it possible that, about 14 billion years ago, time began, that before
then there was no time. In fact, the statement \"before then\" may be
meaningless, since there was no before if there was no time.

Could time stop?

What is the meaning of \"now\"? Can you explain it to someone? Can you
write a paragraph explaining what it means? Does it mean the same thing
to different people? Is my \"now\" the same as your \"now\"? What are
you doing right now?

Why do we remember the past? Could the Universe be such that we would
remember the future instead? Or would we then call it the past? Could we
remember some of the past, and some of the future \-- perhaps a 50/50
mix?

What is the \"pace\" of time? Is the rate of time set by our heartbeats
and our rate of thinking? Is there any meaning to the \"passage of
time\"? Does time \"move\"? If time sped up, and slowed down, would we
notice?

Space exists in three dimensions (at least). Could there be two
dimensions of time? Could time run simultaneously in two separate ways?
Or three? What would life be like with two dimensions of time?

Answers to the Questions

Actually, I don\'t know the answer to any of them. I have some
speculation about remembering the past, rather than the future, but that
is about it. Most of the questions I have given above are not considered
part of physics, but I think that is only because we have made so little
progress in understanding them.

END OF CHAPTER

Quick review

An event can be specified by four dimensions: three positions, and the
time of the event. But the time interval between two events depends on
the frame of reference. The amount of time experienced by a moving
traveler is less than that by someone who is stationary, by the factor

ƒ = sqrt\[1-(v/c)^2^\]

This difference in experienced time has been verified by experiments
with clocks on airplanes and with accelerated radioactive particles. For
low velocities, ƒ ≈ 1, and that is why we don\'t usually notice the
effect. But as v gets close to c, the value of ƒ can be much less
than 1. The same factor also describes length contraction, also known as
\"Lorentz contraction.\" According to this result, objects get shorter
along the direction of motion.

Velocities don\'t add in the usual way. No matter who observes light, it
will appear to be moving at c. That is the \"invariance of the speed of
light.\" If the object is moving less than c, then it will still be
moving less than c for all observers, no matter how fast the observer is
moving.

The energy of a moving object is given by the famous equation E = mc^2^.
This includes both rest energy and kinetic energy. The m in this
equation is the kinetic mass, which grows at high velocity according to
the equation m = m~0~/ƒ = γ m~0~. (m~0~ is a number called the \"rest
mass\" that doesn\'t change.) At low velocities, the total energy is the
rest energy plus (1/2)mv^2^. Rest mass can be turned to kinetic energy.
This is done in nuclear decay, and in annihilation. A photon has rest
mass zero. Such particles do not experience time, and cannot undergo
decay. We used to think neutrinos also have rest mass zero, but since
they can change into other neutrinos, they must have some mass.
Einstein\'s equation E = mc^2^ was not important in the invention of the
nuclear reactor or the atomic bomb. Tachyons are hypothesized particles
that only travel at speeds faster than light. They reach zero energy at
infinite velocity, and infinite energy when they travel at c.

Because of the variability of time in relativity theory, it is
impossible to define \"simultaneous\" in an absolute way, unless two
events occur at the same location.

Essay questions

What usual concepts of time are upset by the theory of relativity? What
do most people accept as obvious, that turns out not to be true?

Energy can be converted to mass, and mass to energy. Describe how. Give
explicit examples.

Physicists often say that people will never be able to travel at the
speed of light. Explain why they believe that.

Describe the peculiar phenomenon of the \"twin effect\" in relativity.

Short questions

Which of the following quantities do not depend on velocity?

> ( ) m
>
> ( ) m~0~
>
> ( ) kinetic energy
>
> ( ) total energy

In the twin effect, the traveling twin is

> ( ) younger
>
> ( ) lighter
>
> ( ) longer
>
> ( ) older

Neutrinos are believed to have mass

> ( ) zero
>
> ( ) small, but not zero
>
> ( ) infinite
>
> ( ) imaginary

When a particle approaches the speed of light, its mass approaches

> ( ) zero
>
> ( ) infinity
>
> ( ) m~0~

In everyday life, we don\'t see the effects of relativity because

> ( ) they only occur when near c
>
> ( ) they are too small to detect easily
>
> ( ) we have become accustomed to them, so we don\'t notice

Tachyons are particles that

> ( ) travel at c
>
> ( ) have zero mass
>
> ( ) travel faster than c

( ) have infinite energy

[^1]: In physics texts, the formula given is usually for gamma rather
    than for ƒ. Gamma = γ = 1/ƒ. Then, for the twin effect, or for the
    Lorentz contraction, you divide by gamma. For this text, it is
    easier to use ƒ, and that way avoid two canceling divisions.

[^2]: Here is a way to do the calculation without a computer. This is
    only for those who love math: when v is very small, we can write
    sqrt\[1 -- (v/c)^2^\] ≈ 1 -- (v/c)^2^/2. (To prove that, multiply 1
    -- (v/c)^2^/2 by itself, and ignore all terms that have (v/c)^4^ in
    them.) In our example, (v/c)^2^ = 8 x10^-13^. So ƒ = 1 - 4 x10^-13^.
    That is the number I gave in the text.

[^3]: I don\'t expect you to read this, but here is the reference:
    Hafele J.C. & Keating R.E. -- Science, 1972, vol. 177, p.166.

[^4]: In their work, Haefele and Keating had to calculate the gravity
    effect -- something we haven\'t discussed yet, as well as the twin
    effect. Their result matched the prediction for the combined
    effects.

[^5]: I\'ve written a more detailed article on this that was published
    in the American Journal of Physics, Vol. 40, pages 966-969.

[^6]: Assume that (v/c) = β is a very small number. There are two key
    approximations you must make. The first is that sqrt(1 - β^2^) ≈ 1
    -- β^2^/2. The second is that 1/(1 -- β^2^/2) ≈ 1 + β^2^/2. Both of
    these approximations can be derived using algebra; calculus is not
    required. (You can check the first equation by squaring it; you can
    check the second by cross-multiplication; in both cases throw away
    the tiny β^4^ terms.)

[^7]: One way to see that is the fact than any number x divided by zero
    is infinite: ∞ = x/0. If you cross multiply, you get 0 ∞ = x

[^8]: Even without invoking the \"zero time\" argument, you can prove
    they don\'t decay from the more formal approach of showing such a
    decay cannot simultaneously conserve energy and momentum.

[^9]: There is an exception. If the two events occur at the same
    location, then the order of events is unambiguous.

[^10]: To be precise, it depends on the velocity of the frame of
    reference, not on the velocity of the observer.

[^11]: The following math is only for those who have studied the Lorentz
    transformation in a different course. Let the two events be (x1, t1)
    and (x2, t2). In the moving frame, they occur spaced by a time
    interval ∆t\' = γ(∆t - u∆x/c^2^) where u is the relative velocities
    of the two frames. ∆t\' will have a different sign from ∆t only if
    ∆x/∆t \> c^2^/u \> c. That is possible if the two events have a
    tachyon moving from one to the other that has ∆x/∆t \> c.