relativity by liwenting

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									                     Relativity

Special Relativity

What does it mean to      If we both measure the
measure something?        same object with the same
                          tools, should we get the
                          same result?

What does it mean to      Should the laws of physics
know something?           be the same for
                          everybody?

What does it mean to      You‘re in a spacecraft and a
be in motion?             comet zips by. Are you
                          moving or is the comet
                          moving?
We have this idea that ―physical reality,‖ whatever that is,
ought to be independent of who/when/where/how a
measurement is made.

Electromagnetic theory was perfected by Maxwell and others
in the late 1800's.

Water waves propagate through water, sound waves
propagate through air.

It is not critical to electromagnetic theory, but it was believed
that electromagnetic waves propagated through the ―ether,‖
relative to some universal reference frame.*
The ether, being ethereal, proved very difficult to detect!


*Imagine the ether attached to this universal reference frame. If you are
moving relative to it, you experience an ―ether drift.‖
Newtonian Relativity Theory                        T1

You swim 200 meters downstream in
a river, turn around, and swim 200
yards upstream. It takes a time T1.

You swim 200 meters perpendicular
to the river bank, turn around, and                         T2
swim 200 yards back. It takes a time
T2 .

T1 and T2 are different. Newtonian
―relativity theory‖ shows you how to             river
calculate T1 and T2.                            current

If you make the ―same‖ measurement on light moving
through the ether, you ought to get the ―same‖ result, T1
and T2 are different.
Michelson and Morley built an interferometer capable of making
such a measurement.
                                     mirror



                                     ―partial‖
                                     mirror
      light                  A
                                                 mirror
      source                     B

Half the light follows path A.
                                                  hypothetical
Half the light follows path B.                    ether drift
                                      detector
The dashed line portions of the paths are oriented differently
relative to the ether drift.
If the times to travel paths A and B are the same, the two light
beams arrive in phase and interfere constructively.

If the times are different, the beams interfere destructively.

Measurement of changes in interference fringe shifts allow
you to deduce the time difference.




                        A
              
                            B

                                  hypothetical
                                  ether drift
But wait! you object. It is impossible to make paths A and B
exactly the same length.

An observed fringe shift might be due to the path length
difference, or it might be due to the different orientations of
the path relative to the ether drift.


So you take a measurement, rotate the apparatus 90
degrees in the horizontal plane, and take another
measurement.

The difference between the two measurements allows you to
very precisely measure the time difference due only to the
ether drift.
Michelson and Morley did the experiment in July, 1887. They
found nothing.

No ether drift.              (Less than 5 km/s; current upper limit is 15 m/s.)


No ether.*

They tried it again later, in case during the July measurement
the earth was coincidentally at rest with respect to the ether.
They got the same results.




             *No universal frame of reference?
I‘ve seen sources that say this result wasn‘t terribly
bothersome, because the ―ether‖ was a conceptual
convenience, and was not required to make E&M theory
work. I‘ve seen other sources that say this was
―devastating‖ at the time. It certainly created a problem.


In fact, if you believe Michelson and Morley and Maxwell,
you are forced to conclude that the speed of E&M radiation
is the same in all non-accelerated reference frames,
regardless of the motion of the radiation source. A
bit difficult to accept!
How do you reconcile the lack of a universal reference
frame with the idea that everybody's measurement of
the same thing ought to produce the same result?

Einstein, 1905, Special Theory of Relativity

Special theory of relativity treats problems involving inertial
(non-accelerated) frames of reference.

We believe the laws of physics work and are the same for
everybody.
Experiment demands that the speed of light be constant.

Let‘s make these two things our postulates and see where we
are led. The validity of the postulates will be demonstrated if
the predictions arising from them are verified by experiment.
Postulates of special theory of relativity:

     the laws of physics are the same in all inertial
    reference frames

     the speed of light in free space has the same value for
    all inertial observers*


The first ―makes sense.‖ The second is required by
experiments but contradicts our intuition and common sense.


*independent of the motion of the source or relative speeds
of observers!
Time Dilation – a consequence of the two assumptions

Let‘s begin with a definition, and then construct a clock.

The time interval between two events which occur at the
same place in an observer‘s frame of reference is called the
proper time of the interval between the events. We use t0
to denote proper time.
Suppose you are timing an event by clicking a stopwatch on at the start
and off at the end. In order for the stopwatch to measure the proper time,
the ―start‖ and ―stop‖ events must occur at the same place in your frame of
reference.

You‘ve been chosen to be a timer at a track meet, so you go stand by the
finish line. You start your stopwatch when you see the puff of smoke from
the starter‘s gun at the starting line, and stop it when the first runner
crosses the finish line. Did you measure the proper time for the sprint?
     Now let‘s make a clock.

                                      mirror




                    L0




                          tick tock
                              
                              laser with built-in light detector

―It's not that I'm so smart , it's just that I stay with problems longer.‖—A.
Einstein
How long does a ―tick-tock‖ take?

                   mirror
                                    time = distance / velocity

                                    t0 = 2L0 / c
    L0

                                       ―I measure proper time
                                       because the light pulse
                                       starts and stops at the
                                            same place.‖
           
           laser
Now put this clock in a transparent spacecraft and observe
as it speeds past.


                                                 ―I don‘t measure
                                              proper time because
                                             ―tick‖ and ―tock‖ occur
                                               in different places.‖

                    L0




   tick                             tock



  entire clock moves with speed v
How long does a ―tick-tock‖ take? Let the total time be t.


                                     distance = velocity · time
                                                          2
                                                t
                                               2
                                      D = 2 L +v 
                                               0
                                                2
                 L0




                                
          vt/2           vt/2


   v
According to the second postulate of special relativity, light
travels at a speed c, so D = ct.

We also know the proper time from our ―stationary clock‖
experiment: t0 = 2L0 / c

                                2
                    t
                   2
          D = 2 L +v 
                   0                        ...(1)
                    2

                       D = ct              ...(2)

                         ct 0
                    L0 =                   ...(3)
                          2
Solving (1) and (2) for t and replacing L0 using (3) gives:

                                    t0
                           t=
                                 1 - v2 c2

Note that (1-v2/c2)1/2 < 1 so t > t0. It takes longer for an
event to happen when it takes place (is timed) in a reference
frame moving relative to the observer than when in takes place
in the observer's reference frame. Time is dilated.* This
applies to all clocks.
Everybody chant: ―a moving clock ticks slower, a moving clock ticks
slower, a moving clock ticks slower…‖ If a moving clock ticks slower, it
counts fewer seconds.

*How to remember what ―dilated‖ means. Pupils in your eye can dilate or
contract. Dilate must be the opposite of contract, so ―dilate‖ must mean
take on a larger value.
A moving clock ticks slower.

If I time an event which starts and stops in in my frame of
reference, I measure t0.

If I use my clock to time the same event as it takes place in a
reference frame moving relative to me, I measure t>t0.

In the latter case, I claim my clock, which measured t, is
correct, so that an identical moving clock, which would
measure t0 in the moving reference frame, is slow.


An event must be specified by stating both its space
and time coordinates.
Example: the Apollo 11 spacecraft that went to the moon
traveled a maximum speed of 10840 m/s. An event observed
by an astronaut in the spacecraft takes an hour. How long
does an earth observer say the event took?

Problem Solving Step 0. Think first!

Always ask: what is the reference frame of the event? Is the
observer in this reference frame or moving relative to it?

The event took place in the spacecraft. The proper time t0 is
the time measured in the spacecraft. Thus, t0 = 3600 s.

The ―observer‖ in this problem is the person on earth, not the
astronaut! The earth observer measures t.
Problem Solving Step 1. Draw (if appropriate) a fully-
labeled* diagram. (*Include values of known
quantities.)

If it helps you to draw a sketch of the earth, a spacecraft, and
a couple of stick figures, do so!

Include values of known quantities. c = 3x108, v = 10840,
t0 = 3600.

If you use SI units throughout, your answer will be in SI units,
and I only need to see units with your final answer.

If you mix systems of units, show the units at each step.*

You can always show units at each step if it helps you.

Sooner or later, if you mix units, you will suffer pain.
Problem Solving Step 2. OSE.


So far, we only have one relativity OSE:
                                    t0
                          t=
                                1 - v2 c2




―Put your hand on a hot stove for a minute, and it seems like an hour. Sit
with a pretty girl for an hour, and it seems like a minute. THAT'S
relativity.‖ – A. Einstein.
Problem Solving Steps 3 and 4. Solve algebraically
first, then substitute values.

The algebra is already done in this case.
                                          3600
                        t=
                               1 - (10840 3×108 )2

                             t = 3600.00000235 s.

Not a big difference, but it is measurable. The actual
experiment has been done with jets flying around the earth,
and the predicted time dilation has been observed.1
As expected, the earth observer measures a bigger number for the time.
The moving clock on the spacecraft measured a smaller number. The
moving clock ticks slower.
1J.   C. Hafele and R. C. Keating, Science 177, 186 (1972).
What if v>c? It can't happen. We‘ll see later that it would
take an infinite amount of energy to accelerate an object up
to the speed of light.

What about time running backwards? Sorry, time always
runs forwards.

What about seeing an event before it happens? Can't,
because c is finite.

However, because of time dilation, events which appear to
be simultaneous in one reference frame may not appear to
be simultaneous in another reference frame.



―The only reason for time is so that everything doesn't happen at once.‖ –
A. Einstein
On the constancy of c…

Recent research suggests that c may not be constant…

Several researchers in Australia have been studying light
absorbed by distant gas clouds about 12 billion years ago.

The fine structure in spectral lines (i.e., the spacing of
multiple lines close together) depends mathematically on the
fine structure constant:
                             e2
                        
                            20hc

In this formula, e is the charge of an electron, 0 is a
constant you encountered in E&M, c is the speed of light,
and h is another constant important in quantum mechanics.
Everything in the formula for  is a constant.

However, the data obtained by the Australian group suggest
that  had a larger value when the light they observed was
emitted than it does now.
                            2        seems to have changed by
                        e           0.001% in 12000000000 years.
                                  That‘s a change of
                       20hc
                                    0.00000000000008% per year.

If  has been increasing over time, then either the charge
on an electron has been increasing, or 0, h, or c have been
decreasing.

According to a commentary* put out by the American
Physical Society…

*So it is an admittedly biased opinion of the commentary author.
―Since the effect on the laws of physics of increasing the
electronic charge are too awful to contemplate, they figure
light is going slower. That kills relativity, but my mail
indicates nobody but physicists believe that stuff anyway.‖—
Bob Park

The Australian researchers have reported on this work three
times in recent years, including 2001 in Physical Review
Letters* and in 2002 in Nature**. It seems like I am
constantly hearing ―new‖ reports of findings that c is
decreasing, but it is all essentially this one group reporting
their work as it progresses.



*Arguably the most prestigious US physics journal.
**The most prestigious science journal known to man.
To date no one else has reproduced this result.

Some possibilities:
   The Australians* could have made a mistake (unlikely).
   Their results could be a statistical fluke (unlikely).
   A yet-undiscovered systematic error could have
  influenced their results.
   The interpretation could be wrong.
   They are correct.
   ???

This is important enough that others will be investigating
carefully. We should know the results within a few years.

*I‘m not picking on Australians. They are as smart as we are. I am
using the country of origin as a convenient way to identify the research
group.
What if they are right?

Theories you will learn in this class will supersede theories
you learned earlier (e.g., Newtonian Mechanics).

You should not think of the earlier theories as being ―wrong.‖
Rather, the new theories are better, and incorporate the old
ones within them.

Relativistic mechanics reduces to Newtonian mechanics in
the limit of small relative velocities.

Use Newtonian mechanics when the error introduced is
small (because it is easier to use). Use relativity only if you
must!
"If we knew what it was we were doing, it would not be called research, would
it?“—A. Einstein
If they are right…

 There will be profound implications for cosmological
theories.

 Someone will have to re-think special relativity. Someone
will have to come up with a new theory which incorporates
all of special relativity but goes beyond it to include the
slowly-changing value of c.

 This may have profound implications for mankind (as did
special relativity). It may not. We‘ll see.

 Newtonian mechanics will still work just fine as long as
velocities are not too big.

 Lots of physicists will have nice jobs for a long time to
come.        And now, back to our show…
Let‘s consider another problem that time dilation helps us
solve.

Has anyone here ever felt a muon?

Does anybody even know what a muon is?

A muon is an elementary particle with a mass 207 times that
of an electron, and a charge of either +e or –e. Muons are
created in abundance at altitudes of 6 km* or more when
cosmic rays collide with nuclei in the atmosphere.

Fortunately, muons interact only very weakly with matter,
which is why it is OK that many of them are passing through
your body right now.

*This is in the upper reaches of the troposphere, the part of the atmosphere in
which we live.
Muons travel with speeds of about 0.998 c (fast!) and have
an average lifetime of 2.2 s (2.2x10-6 s).

How far can an average muon travel during its lifetime?

                                d=vt
                d = 0.998 · 3·108 · 2.2·10-6 = 0.66 km.

How can muons get through the 6 or more kilometers
of atmosphere between their birthplace and us if they
only live long enough to travel 0.66 km?



OK, a some will go more than 0.66 km, and some less, but not many, and not by much. So
the question stands.
Time dilation!
I say the muon‘s clock ticks slow. I say that while the muon
thinks* its clock ticks 2.2 s, I observe that it actually
ticks
                                2.2 ×10-6
                        t=
                               1 - (0.998)2
                            t  34.8s
During this time the muon travels a distance

             d = 0.998 · 3·108 · 34.8·10-6 = 10.4 km,

so the average muon will reach me before decaying.

*Of course, a muon doesn‘t ―think‖ anything, but we use words like that to help
us form a mental image of the process. If you prefer, imagine a nano-human
riding on the muon and reporting what he/she sees.
Double-check: what is the event, who is the observer, and
who measures the proper time.


The event is the muon ―living.‖


The event does not take place at a single location in my
reference frame, so I measure the dilated time, and the
calculation was correct.
One important aspect of relativity is that there is only one
reality. If I see the muon arrive at the surface of the earth,
the muon must agree that it actually did arrive at the
surface of the earth.
Our average muon ―says‖ there is no doubt whatsoever that its
lifetime is 2.2 s, and during that time it travels 0.66 km. I say
the muon reaches the surface of the earth. The muon
says it doesn’t??

―I thought you said time dilation would help us solve the muon
problem.‖ We seem to have created a new problem.

Either we have encountered two different realities, or else there
is…

 ―Relativity teaches us the connection between the different descriptions
 of one and the same reality.‖—A. Einstein
Length Contraction
―The faster you go, the shorter you are.‖—A. Einstein

If two observers in relative motion measure different times for
an identical event, what makes us think they should measure
the same lengths for an identical object?

The formula for length contraction is not terribly difficult to
derive. I‘ll lend you a book if you are curious. Here is the
formula.
                 OSE:      L  L0 1  v 2/c2
The Proper Length, L0, of an object is its length as measured
in its own rest frame.

An observer measuring the length of an object moving relative
to him will measure a length L less than the length L0 he
would measure if he were not moving relative to the object.
Let me demonstrate length contraction using a meter stick…
The length contraction occurs only along the direction of
relative motion. A spacecraft moving past an observer at
nearly the speed of light will seem to be very short in length
and normal diameter.
A muon created at an altitude of 10.4 km would say that
during its lifetime it saw an atmosphere of length

      L  L0 1  v 2/c2  (10.4)  1  0.9982  0.66 km
I say the muon gets to earth because its lifetime is longer. The
muon says it gets to earth because the atmosphere is shorter.
Different descriptions of the same reality.

Be careful when you talk about the lifetime of a particle
moving with v close to c. You need to specify the reference
frame in which the lifetime is measured!
The Twin Paradox

A and B are 20 year old twins. A travels on a spaceship at v =
0.8c to a star 20 light years* away and returns.

B, left behind on earth, says the trip takes 2·20/0.8 = 50 years.
B is 70 years old when A returns.

B also observes that A‘s clock (which is identical to B‘s) ticks
slowly, and records less time. If the event in question is the
ticking of A‘s clock, then the 50 years calculated above is the
dilated time t (why?).

*A light year, y, is the distance light travels in one year. Thus, y = (1 year)·(c). If D is
a distance expressed in light years, then the number of years it takes to travel that
distance at a speed of v is found from time = (distance) / velocity. Thus:
    time in years = (distance in light years) / (velocity expressed as a fraction of c).
The proper time, which in this case is amount of time
recorded by a clock in the spacecraft, is is found by solving
our time OSE for t0:
                       t0 = t 1 - v 2 c2

                      t 0  50 1  0.82
                          t 0  30
According to B (who was left back on earth), A‘s clock only
ticked 30 years, so that A is 20 + 30 = 50 years old on
return to earth.

At the end of the trip, B, left behind, is 70 years old. A, who
made the trip, is 50 years old. Can this be possible?

Yes! Absolutely! and it was verified experimentally in the
jets-around-the-world experiment mentioned earlier.
Now here‘s the paradox. A moving clock ticks slower. This
applies to all observers. A, on the spacecraft, sees B move
away and then come back.* A says B‘s clock ticks slower. A
does the calculation presented on the last slide and concludes
that at the end of the trip, B is 50 and A is 70.


That‘s the famous twin paradox. It would appear that each
twin rightfully claims the other aged less. Have we
discovered an example of the existence of two
different, mutually exclusive realities?




*Remember, there is no absolute reference frame for specifying motion. Motion is
relative! An observer is free to say ―I am at rest; you are the one moving!‖
When you encounter a paradox like this you can be sure that
someone has pulled a fast one on you.
In this case, an unwarranted calculation was made.

Special relativity applies only to observers in inertial (non-
accelerated) reference frames. A had to accelerate (very
rapidly) to leave earth and get up to speed, and again when
turning around to head home, and a third time when landing
on earth.*

A is not allowed to use the equations of special
relativity! B is, and B‘s calculation is correct: A comes back
20 years younger.
If you examine the problem carefully, it‘s only the turning
around part that causes A trouble.

What‘s poor A to do? Doesn‘t a moving clock tick slower?
Yes, so evidently during A‘s period of extreme acceleration, B‘s
clock (as observed by A) would tick incredibly fast. Isn‘t A
allowed to use the laws of physics? Yes, but it would have to
be general relativity.
We won’t have completely eliminated the paradox
unless we can find a description for A’s reality that
agrees with B’s reality.

A, in the spacecraft, needs to reconsider the distance traveled.
During the ―out‖ portion of the trip, A will say that the actual
distance traveled was

  L  L0 1  v 2 c2  20 1  0.82  12 light years,

and that the back portion was also 12 light years. 24 light
years at a speed of 0.8 c takes 30 years so A ages 30 years
during the trip, and comes back at age 50.
B tells A ―you are younger because your clock ticked slower.‖

A says ―I am younger because the trip covered less distance
than you thought.‖

Same reality, two different descriptions.

There are a number of famous paradoxes based on relativistic
calculations. Typically, someone makes an invalid calculation
(usually on purpose, to see if they can trick you).



In another paradox, where a very fast runner tries to put a 10 meter pole
in a 5 meter barn, a paradox arises because… (I’ll let you ponder that
and come back to it later).
Electricity and Magnetism

The material we‘ve been studying is fascinating and thought-
provoking, but it is not how Einstein‘s theory of relativity came
into being.

―What led me more or less directly to the special theory of
relativity was the conviction that the electromagnetic force acting
on a body in motion in a magnetic field was nothing else but an
electric field.‖—A. Einstein.


In other words, Einstein believed that what you and I might
call a magnetic force is really just an electric force in another
inertial reference frame.
Consider a conducting wire and a positive test charge.



       +       -    +                  +   -       +
  -                                -                   -
               +    -      +               +   -
   -       -
       +            +          -       +   -       +




                           +

What force does the test charge ―feel‖ due to the charges in
the wire?
    Repulsion, because there is a – closest to the test +?

            No net charge inside the conductor.


       +         -    +                  +   -        +
-                                    -                    -
                 +    -      +               +    -
-            -
       +              +          -       +   -        +


           No electric field outside the conductor.

                             +

                          No force!
What does the test charge see when an electric field is
applied and current flows?
                             E

                      -               -               -                   -
          +       -           +                   +               +
    -                     -                   -           -
          -                                       -                   -
                  +           -       +               +       -
     -`       -                   -                                   -
          +           -       +           -       +   -           +




                                          +
The test charge ―observes‖ that the space between the moving
electrons is contracted. There are more electrons in the part
of the conductor nearest the test charge!
The test charge ―observes‖ that the moving electrons are
closer together than the stationary protons, and therefore
"feels" a Coulomb attraction.

A human observer is unable to see the electrons, and
attributes the attraction to a ―magnetic force‖ generated by
the moving charges.

Same reality, two different descriptions!

And both descriptions are incomplete or mildly
troublesome, as we will see shortly…

Beiser‘s presentation of this material is different, but
equivalent.
If you think about it, this presentation is bothersome.

To illustrate, I need to talk about conservation and invariance.
A quantity is relativistically invariant if it has the same value in
all inertial frames of reference.

           The speed of light is relativistically invariant.
           Time is not relativistically invariant.
           Length is not relativistically invariant.
           Electric charge is relativistically invariant.*
*All observers agree on the total amount of charge in
a system.
A quantity is conserved if it has the same value before and
after some event. Don‘t confuse conservation with invariance.
It is a fact that electric charge is both conserved and
relativistically invariant.

Our thought experiment with the conductor and test charge
suggests that a conductor which is electrically neutral in one
reference frame might not be electrically neutral in another.
How can we reconcile this with charge invariance?
My modern physics textbook author claims there is no
problem, because you have to consider the entire circuit.
Current in one part of the circuit will be balanced by opposite
current in another part.
Although the explanation is correct, I don‘t find it satisfying.*
Maybe the pole-in-barn paradox will help us understand.

*It seems logical that if moving electrons are closer in one part of the
circuit, they ought to be closer in other parts of the circuit too, so that
the conductor is no longer neutral and charge is not conserved.
The Pole-Barn Paradox

A speedy runner carrying a 10-
meter pole approaches a barn
that is 5 meters long (short
barn!), with open doors at each
end. A farmer stands nearby,
where he can see both front and
back door at the same time.

a) How fast does the runner have to go for the farmer to
observe that the pole fits entirely in the barn?

b) What will the runner observe?
The answer to a) involves a simple length contraction*
calculation.
For the pole to fit in the barn, the farmer must measure a
contracted length L = 5 m for the pole of proper length L0 =
10 m.
                     L  L0 1  v 2 /c2

                      5  10 1  v2 /c2

The result is v = 0.866 c. If the runner is going that fast, or
faster, the farmer observes the pole to fit inside the barn.


Length contraction is often called the Lorentz contraction, named after
the scientist who discovered the mathematical transformations which
lead to the equation for length contraction.
The answer to b) starts with another length contraction
calculation.
The runner is moving …

no, the runner isn‘t moving. The runner sees the barn moving
towards him at a speed of v = 0.866 c.

The runner says the speeding barn has a length equal to

                  L  L0 1  v 2 /c2

                  L  5 1  0.8662
                      L  2.5 m.

The pole can’t possibly fit inside the barn.
How do we explain this paradox? Which observation is      the
physical reality?

The answer: both observations are correct!

A detailed calculation (I can lend you the book it is in, if you
are interested) shows that the runner observes the rear end
of the barn arriving* at the front end of the pole long before
the front end of the barn arrives at the rear end of the pole.
The pole doesn’t fit!
Events which are simultaneous in the farmer’s frame
of reference (front pole arriving at back barn and back
pole arriving at front barn) are not simultaneous in the
runner’s frame of reference.
*Remember, the runner sees the barn moving past him.
Simultaneity is not a ―universal physical reality.‖*

Now I‘m no longer worried about the test-charge-plus-
conductor example. At a certain instant in time I may observe
an excess of moving negative charge in the portion of the
circuit nearest me, but does not mean I can claim there is a
net excess of moving negative charge in the entire circuit at
that instant in time.

Now where were we before this interruption started…



―Because simultaneity is a relative concept and not an absolute
one, physical theories that require simultaneity in events at
different locations cannot be valid.‖—Beiser, Modern Physics.
An observer who doesn‘t know about relativity, or even one
who knows about relativity but invokes charge invariance, will
claim that the conductor has a neutral charge density and
invents a ―magnetic‖ force to explain the attraction.


But the ―magnetic‖ force is present only when current is
flowing. It is not valid to talk about a separate ―magnetic‖
force. You must talk about the ―electromagnetic‖ force.


What you call ―magnetic‖ force is just a manifestation of the
Lorentz contraction and Coulomb‘s law, and is not a separate
force of nature.
The mathematical transformations which lead to our
relativistic equations for length and time were derived by
Lorentz to make Maxwell‘s equations invariant in inertial
reference frames.*

Because Maxwell‘s equations are invariant in inertial reference
frames, special relativity does not demand that we correct
them.
On the other hand, when it comes to Newton‘s Laws…




*Part of Einstein’s genius was realizing that Lorentz
was on to something big!
1.7 Relativistic Momentum

We believe very strongly that momentum is conserved. Let‘s
see what effect a relativistic calculation has on momentum.

Here‘s the essence of the calculation
my text uses:                                     B

Take two observers with identical
(including mass), elastic balls
(elastic, so that kinetic energy is
conserved).

Have the observers stand along the
y-axis, equal distances away from
the origin, and throw the balls with
equal speeds (call them VA and VB′)               A
towards the origin.
There is no new or exciting physics here. Using conservation
of momentum, you could easily show that the two balls have
equal and opposite velocities after the collision.


Now, just for kicks,
lets put one of the            S′ throws B
observers in motion,
with a speed v in the
+x direction. Call that
observer S′. To the               v
other observer, S, this
is what the collision
looks like.


                                             S throws A
 Let the speeds of the balls as measured by S be VA and VB
 and let the y-component of the (identical) distance each one
 travels be Y/2.*
                                                   S′ throws B
 The travel time (to collision and
 back) for ball A as measured by
 observer S is T0.

 The travel time which observer                        v
 S measures for ball B is the                Y
 dilated time T.



                                                                   S throws A

*The y-components of the distances are identical because v has no y-component.
                                       Y          Y                   Y
According to observer             T0          T                VA 
                                       VA         VB                  T0
S:

                                       Y
According to observer             T0 
                                        
                                       VB
S′:
                            T0
Time dilation:     T
                          1- v2
                                  c2

                                                Y 1- v2
                                           Y
                                       VB               c2
According to observer S, VB
                                           T         T0
is:

The two boxed equations give the speeds S observes for
balls A and B.
VB<VA. Same distance, S says B takes longer so B moves slower.
If we use the classical (Newtonian) definition for momentum,
S says that
                                                   Y 1- v2 2
                       Y                                  c
    p A  m A VA  m A          p B  m B VB  m B
                       T0                             T0
       If mA=mB then pB<pA, as expected -- remember, VB<VA.

If mA and mB are identical, then momentum is not conserved.


This analysis is correct, but I find it confusing because details
are left out.
                                                     If we could somehow
        red—momentum of B                            “make” ball B have more
   “purple”—momentum of A                            momentum, then momentum
                                                     would be conserved.
      green—total momentum
            (not conserved)
                                 Before After
Non-conservation of momentum is an alarming idea. What
can we do to fix this situation?

 “Making” ball B have more mass would conserve momentum!


              mA
If m B                     then the problem would be ―fixed.‖ Kind of.
                   2
            1- v
                       c2
I say ―kind of‖ because both balls A and B would be moving.
We would really like to compare the mass of a moving object
with the mass of an identical, non-moving object.

If we let VA0 then we have the proper condition for
comparison, and you can show that,

                               m
                    m(v)                    ,
                                    2
                             1- v
                                        c2

where m is the mass of the ball at rest, and m(v) is the mass it
needs to have when it is moving, if we believe in conservation
of momentum.
In the old days* we then said ―OK, m is the mass of the
object at rest and m(v) is its mass when it is moving. Let‘s
call m0 the rest mass and m the mass when it is moving
(‗relativistic mass‘).‖ This notation is consistent with our
equations for time dilation and length contraction, so we
have
                              m0
                        m           .    Not an OSE this
                                v2 2      semester. Don’t use it!
                             1-
                                  c


This was Einstein‘s original approach, but later he said it is
―not good.‖




*When I studied relativity in college, and in the previous edition of our text.
The ―new‖ approach is to say ―Look, mass is mass. We
believe it is something fundamental. If we believe in
conservation of momentum, we had better change our
definition of momentum.‖
                                                  1
If we define momentum as p  γmv where γ             2
                                                          ,
                                                 1- v 2
                                                        c

Then ―mass is mass,‖ momentum is conserved in our thought
experiment (and in real life), and relativistic momentum
reduces to classical (Newtonian) momentum in the limit
v0.*




*More satisfying than saying “mass changes with velocity.”
So in this most recent version of our text, we are always
going to use the symbol m for mass. It‘s what we called
―proper mass‖ or ―rest mass‖ in the old days.

In the old days, rest mass was relativistically invariant. Now
mass is relativistically invariant. Same reality, just different
use of words.
Let‘s make this new notation official.
                                  1
              OSE :        γ                   ,
                                       2
                                1- v
                                           c2


                                
              OSE :        p  γmv



A consequence of this new definition of momentum:

               dp                        
                                        dp d
                                                 
              F     ma               F     γmv 
                 dt                       dt dt
More consequences of this new definition of momentum:


          
        F v2
     a  1-
        m    c2
                  
                  3/ 2


                                   relativistic momentum
                                   increases without
                                   bound as vc
    For finite F,
    a0 as vc.*

    No finite force
    can accelerate
    an object
    having                                     no limit to classical velocity

    nonzero mass
    up to the
    speed of light!
When can I use “rest” mass, and when do I have to use
                            m
                m(v)                     ?
                                 2
                          1- v
                                     c2


    Object         v                  v/c      m(v)/m
     jogger     10 km/h     .000000009           ≈1
      space     104 m/s          0.000033     1.0000001
     shuttle
    electron    106 m/s              0.0033     1.001
    electron    108 m/s              0.333      1.061
1.8 Mass and Energy

From your first-semester physics course:

                         dγmv
          KE   Fds          ds.
                           dt

Use the definition of  and integrate by parts to get

          KE  γmc 2  mc 2  γ  1mc 2

                γmc 2  mc 2  KE.

Assuming potential energy is zero (we can always choose
coordinates to do this), we interpret mc2 as total energy.

                  E  mc2  KE.         The “box” indicates an OSE.
When an object is at rest KE = 0, and any energy that
remains is interpreted as the object‘s rest energy E0.

                           E 0  mc 2 .

When an object is moving, its total energy is

                                        mc2
                   E  γmc2                        .       This is really just a
                                      1 v2                 variation of the OSE
                                               c2           on the previous slide.




         This is the closest you’ll come to seeing E=mc2 in this class.
         In the “old days,” E=mc2 would have been written E=mc2.
These equations have a number of interesting implications.

Mass and energy are two different aspects of the same “thing.”


Conservation of energy is actually conservation of mass-energy.


The c2 in E0=mc2 means a little mass is “worth” a lot of energy.


Your lunch: an example of relativity at work in “everyday life.”



          Total energy is conserved but not relativistically invariant.
               Rest (or proper) mass is relativistically invariant.
        Mass is not conserved! (But it is for the purposes of chemistry.)
Example: when 1 kg (how much is that?) of dynamite explodes, it
releases 5.4x106 joules of energy. How much mass disappears?

This is actually a conservation of mass-energy problem. If
this material were presented in Physics 23, I would make
you start with your conservation of mass-energy OSE and
derive the appropriate equations from there.

For Physics 107, it is sufficient to realize that the problem is
just asking ―what is the mass equivalent of 5.4x106 joules of
energy?‖
          E 0  mc 2
              E0
           m 2
              c
              5.4 106
           m            6 1011 kg.    Conservation of mass is a
               
               3 108 2
                                         very good approximation!
If we are to claim relativistic mechanics as a replacement
theory for Newtonian mechanics, then relativistic mechanics
had better reduce to Newtonian mechanics in the limit of
small relative velocities.
                                mc2
         KE  γmc2  mc2                  - mc2 .
                             1 v2
                                      c2

Our text shows that for v<<c,

                         1
                     KE  mv2 .
                         2
When can I get away with using KE = mv2/2, and when do I have
to use KE = mc2 - mc2?

Use Newtonian KE every time you can get away with it!
Use relativistic KE only when you must!

If v = 1x107 m/s (fast!) then mv2/2 is off by only 0.08%.
Probably OK to use mv2/2. If v = 0.5 c, then mv2/2 is off by
19%. Better use relativity.
Energy and Momentum
Total energy and magnitude of momentum are given by

             mc2                                             mv
       E                                        p                      .
                   2                                            2
            1- v           2                             1- v
                       c                                            c2

With a bit of algebra, you can show

                               E  p c  mc
                                2   2 2
                                                 2 2
                                                         .

The quantities on the LHS and RHS of the above equation
are relativistically invariant (same for all inertial observers).
Rearranging:
                                2
                                     
                               E  mc     2 2
                                                 p 2c 2 .
               mc2                      mv
         E                       p                   .
                     2                        2
              1- v           2         1- v
                         c                        c2

Is it possible for a particle to have no mass? If m = 0, what
are E and p?

For a particle with m = 0 and v < c, then E=0 and p=0. A “non-
particle.” No such particle.

But if m = 0 and v = c, then the two equations above are
indeterminate. We can’t say one way or the other.
If m = 0 and v = c, we must use E 2  mc2   p2c2 .
                                                           2




The energy of such a particle is E = pc. We could detect this
particle! It could exist.

Do you know of any massless particles?
            photon
            neutrino*
            graviton**              graviton is to gravity as photon is to E&M field




*Maybe. Nobel prize for you if you show mneutrino = 0.
**Maybe. Nobel prize for its discoverer. Problem: gravitational fields much,
much weaker than E&M fields.
Looking ahead…

Particles having KE >> E0 (or pc >> mc2) become more
photon-like and behave more like waves.

The momentum carried by massless particles is nonzero (E
= pc).

Could you stop a freight train with a flashlight?


Could you stop a beam of atoms with a laser beam?
A note on units.

We will use the electron volt (eV) as an energy unit
throughout this course.
                                
           1 eV  1.6 10 19 C  1 V   1.6 10 19 J

Variations on the eV:
                   1 meV = 10-3 eV (milli)
                    1 keV = 103 eV (kilo)
                   1 MeV = 106 eV (mega)
                    1 GeV = 109 eV (giga)

Because mass and energy are convenient, we sometimes
write masses in ―energy units.‖
An electron has a rest mass of 9.11x10-31 kg. If you plug
that mass into E0 = mc2, you get an energy of 511,000 eV, or
511 keV, or 0.511 MeV.

We sometimes write the electron mass as 0.511 MeV/c2.

It is also possible to express momentum in ―energy units.‖
An electron might have a momentum of 0.3 MeV/c.

If you are making a calculation with an equation like
                     2
                          
                    E  mc  2 2
                                   p 2c 2
and you want to use 0.511 MeV/c2 for the electron mass,
please do. It often simplifies the calculation. But watch
out…
What is the total energy of an electron that has a momentum
of 1.0 MeV/c?
                   E  mc   p2c2
                    2     2 2

                                   2            2
               0.511MeV 2   1.0 MeV  2
           E 
            2
                        c           c
                                    
                      2
                    c              c

              E 2  0.511 MeV   1.0 MeV 
                               2            2




                         
                    E 2  1.26 MeV 2   
                     E  1.12 MeV

Notice the convenient cancellation of the c‘s in the 2nd step.

Avoid the common mistake: don’t divide by an extra c2 or
multiply by an extra c2 in the 2nd step.
General Relativity


Something to think about. Is the mass that goes in F =
ma (or the relativistic version) the same ―thing‖ as the
mass that goes in F = Gm1m2/r2?

Not necessarily!

Experimentally, the two ―kinds‖ of mass are the same to
within better than one part in 1012, and must of us believe
they are the same anyway, so…
“An observer in a closed laboratory cannot distinguish between
the effects of a gravitational field or an acceleration of the lab.”


                    The principle of equivalence.
The principle of equivalence leads one to conclude that light
must be deflected by a gravitational field.

Experimental observation of this effect in 1919 was one of
Einstein‘s great triumphs.

We investigate more about light and gravity in modern
physics classes.




      "If A equals success, then the formula is: A=X+Y+Z. X is work. Y
      is play. Z is keep your mouth shut.“—A. Einstein
More Einstein quotes:
“As far as the laws of mathematics refer to reality, they are not certain, and as far as they are
certain, they do not refer to reality."
"Relativity teaches us the connection between the different descriptions of one and the same
reality".
"I sometimes ask myself how it came about that I was the one to develop the theory of relativity.
The reason, I think, is that a normal adult never stops to think about problems of space and time.
These are things which he has thought about as a child. But my intellectual development was
retarded, as a result of which I began to wonder about space and time only when I had already
grown up."
"The secret to creativity is knowing how to hide your sources."
"The important thing is not to stop questioning.
"Only two things are infinite, the universe and human stupidity, and I'm not sure about the
former."
"Things should be made as simple as possible, but not any simpler."
"Sometimes one pays most for the things one gets for nothing."
"Common sense is the collection of prejudices acquired by age 18.
"Strange is our Situation Here Upon Earth"
"If you are out to describe the truth, leave elegance to the tailor."
"I never think of the future. It comes soon enough.“
"Not everything that counts can be counted, and not everything that can be counted counts."
"The faster you go, the shorter you are."
"The wireless telegraph is not difficult to understand. The ordinary telegraph is like a very long
cat. You pull the tail in New York, and it meows in Los Angeles. The wireless is the same, only
without the cat. "

								
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