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               FREQUENTLY ASKED QUESTIONS ON SCI.PHYSICS - Part 4/4
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Item 24. Special Relativistic Paradoxes - part (a) 

The Barn and the Pole                   updated 4-AUG-1992 by SIC
---------------------                   original by Robert Firth

	These are the props.  You own a barn, 40m long, with automatic
doors at either end, that can be opened and closed simultaneously by a
switch. You also have a pole, 80m long, which of course won't fit in the
barn. 

	Now someone takes the pole and tries to run (at nearly the speed of
light) through the barn with the pole horizontal.  Special Relativity (SR)
says that a moving object is contracted in the direction of motion: this is
called the Lorentz Contraction.  So, if the pole is set in motion
lengthwise, then it will contract in the reference frame of a stationary
observer. 

	You are that observer, sitting on the barn roof.  You see the pole
coming towards you, and it has contracted to a bit less than 40m. So, as
the pole passes through the barn, there is an instant when it is completely
within the barn.  At that instant, you close both doors.  Of course, you
open them again pretty quickly, but at least momentarily you had the
contracted pole shut up in your barn.  The runner emerges from the far door
unscathed. 

	But consider the problem from the point of view of the runner.  She
will regard the pole as stationary, and the barn as approaching at high
speed. In this reference frame, the pole is still 80m long, and the barn
is less than 20 meters long.  Surely the runner is in trouble if the doors 
close while she is inside.  The pole is sure to get caught. 

	Well does the pole get caught in the door or doesn't it?  You can't
have it both ways.  This is the "Barn-pole paradox."  The answer is buried
in the misuse of the word "simultaneously" back in the first sentence of
the story.  In SR, that events separated in space that appear simultaneous
in one frame of reference need not appear simultaneous in another frame of
reference. The closing doors are two such separate events. 

	SR explains that the two doors are never closed at the same time in
the runner's frame of reference.  So there is always room for the pole.  In
fact, the Lorentz transformation for time is t'=(t-v*x/c^2)/sqrt(1-v^2/c^2).
It's the v*x term in the numerator that causes the mischief here.  In the
runner's frame the further event (larger x) happens earlier.  The far door 
is closed first.  It opens before she gets there, and the near door closes 
behind her. Safe again - either way you look at it, provided you remember 
that simultaneity is not a constant of physics. 

References:  Taylor and Wheeler's _Spacetime Physics_ is the classic. 
Feynman's _Lectures_ are interesting as well.

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Item 24. Special Relativistic Paradoxes - part (b) 

The Twin Paradox                                updated 04-MAR-1994 by SIC
----------------                                original by Kurt Sonnenmoser

A Short Story about Space Travel:

	Two twins, conveniently named A and B, both know the rules of
Special Relativity.  One of them, B, decides to travel out into space with
a velocity near the speed of light for a time T, after which she returns to
Earth. Meanwhile, her boring sister A sits at home posting to Usenet all
day.  When B finally comes home, what do the two sisters find?  Special
Relativity (SR) tells A that time was slowed down for the relativistic
sister, B, so that upon her return to Earth, she knows that B will be
younger than she is, which she suspects was the the ulterior motive of the
trip from the start. 

	But B sees things differently.  She took the trip just to get away 
>from the conspiracy theorists on Usenet, knowing full well that from her 
point of view, sitting in the spaceship, it would be her sister, A, who 
was travelling ultrarelativistically for the whole time, so that she would 
arrive home to find that A was much younger than she was.  Unfortunate, but 
worth it just to get away for a while. 

	What are we to conclude?  Which twin is really younger?  How can SR
give two answers to the same question?  How do we avoid this apparent
paradox? Maybe twinning is not allowed in SR?  Read on. 

Paradox Resolved:

	Much of the confusion surrounding the so-called Twin Paradox
originates from the attempts to put the two twins into different frames ---
without the useful concept of the proper time of a moving body. 

	SR offers a conceptually very clear treatment of this problem.
First chose _one_ specific inertial frame of reference; let's call it S.
Second define the paths that A and B take, their so-called world lines. As
an example, take (ct,0,0,0) as representing the world line of A, and
(ct,f(t),0,0) as representing the world line of B (assuming that the the
rest frame of the Earth was inertial). The meaning of the above notation is
that at time t, A is at the spatial location (x1,x2,x3)=(0,0,0) and B is at
(x1,x2,x3)=(f(t),0,0) --- always with respect to S. 

	Let us now assume that A and B are at the same place at the time t1
and again at a later time t2, and that they both carry high-quality clocks
which indicate zero at time t1. High quality in this context means that the
precision of the clock is independent of acceleration. [In principle, a
bunch of muons provides such a device (unit of time: half-life of their
decay).] 

	The correct expression for the time T such a clock will indicate at
time t2 is the following [the second form is slightly less general than the
first, but it's the good one for actual calculations]: 

            t2          t2      _______________ 
            /           /      /             2 |
      T  =  | d\tau  =  | dt \/  1 - [v(t)/c]              (1)
            /           /
          t1          t1

where d\tau is the so-called proper-time interval, defined by

              2         2      2      2      2
     (c d\tau)  = (c dt)  - dx1  - dx2  - dx3 .

Furthermore,
                   d                          d
           v(t) = -- (x1(t), x2(t), x3(t)) = -- x(t)
                  dt                         dt

is the velocity vector of the moving object. The physical interpretation
of the proper-time interval, namely that it is the amount the clock time
will advance if the clock moves by dx during dt, arises from considering
the inertial frame in which the clock is at rest at time t --- its
so-called momentary rest frame (see the literature cited below). [Notice
that this argument is only of heuristic value, since one has to assume
that the absolute value of the acceleration has no effect. The ultimate
justification of this interpretation must come from experiment.]

	The integral in (1) can be difficult to evaluate, but certain
important facts are immediately obvious. If the object is at rest with
respect to S, one trivially obtains T = t2-t1. In all other cases, T must
be strictly smaller than t2-t1, since the integrand is always less than or
equal to unity. Conclusion: the traveling twin is younger. Furthermore, if
she moves with constant velocity v most of the time (periods of
acceleration short compared to the duration of the whole trip), T will
approximately be given by      ____________                              
                              /          2 | 
                    (t2-t1) \/  1 - [v/c]    .             (2)

The last expression is exact for a round trip (e.g. a circle) with constant
velocity v. [At the times t1 and t2, twin B flies past twin A and they
compare their clocks.] 

	Now the big deal with SR, in the present context, is that T (or
d\tau, respectively) is a so-called Lorentz scalar. In other words, its
value does not depend on the choice of S. If we Lorentz transform the
coordinates of the world lines of the twins to another inertial frame S',
we will get the same result for T in S' as in S. This is a mathematical
fact. It shows that the situation of the traveling twins cannot possibly
lead to a paradox _within_ the framework of SR. It could at most be in
conflict with experimental results, which is also not the case. 

	Of course the situation of the two twins is not symmetric, although
one might be tempted by expression (2) to think the opposite. Twin A is
at rest in one and the same inertial frame for all times, whereas twin B
is not.  [Formula (1) does not hold in an accelerated frame.]  This breaks 
the apparent symmetry of the two situations, and provides the clearest
nonmathematical hint that one twin will in fact be younger than the other
at the end of the trip.  To figure out *which* twin is the younger one, use
the formulae above in a frame in which they are valid, and you will find
that B is in fact younger, despite her expectations. 

	It is sometimes claimed that one has to resort to General
Relativity in order to "resolve" the Twin "Paradox". This is not true. In
flat, or nearly flat, space-time (no strong gravity), SR is completely
sufficient, and it has also no problem with world lines corresponding to
accelerated motion. 

References: 
        Taylor and Wheeler, _Spacetime Physics_  (An *excellent* discussion)
        Goldstein, _Classical Mechanics_, 2nd edition, Chap.7 (for a good 
        general discussion of Lorentz transformations and other SR basics.) 

********************************************************************************
Item 24. Special Relativistic Paradoxes - part (c) 

The Superluminal Scissors                           updated 31-MAR-1993 
-------------------------                           original by Scott I.Chase


	A Gedankenexperiment:

	Imagine a huge pair of scissors, with blades one light-year long.
The handle is only about two feet long, creating a huge lever arm,
initially open by a few degrees.  Then you suddenly close the scissors. 
This action takes about a tenth of a second.  Doesn't the contact point
where the two blades touch move down the blades *much* faster than the
speed of light? After all, the scissors close in a tenth of a second, but
the blades are a light-year long.  That seems to mean that the contact
point has moved down the blades at the remarkable speed of 10 light-years
per second.  This is more than 10^8 times the speed of light!  But this
seems to violate the most important rule of Special Relativity - no signal
can travel faster than the speed of light.  What's going on here? 

	Explanation:

	We have mistakenly assumed that the scissors do in fact close when
you close the handle.  But, in fact, according to Special Relativity, this
is not at all what happens.  What *does* happen is that the blades of the
scissors flex.  No matter what material you use for the scissors, SR sets a
theoretical upper limit to the rigidity of the material.  In short, when
you close the scissors, they bend. 

	The point at which the blades bend propagates down the blade at
some speed less than the speed of light.  On the near side of this point,
the scissors are closed.  On the far side of this point, the scissors
remain open.  You have, in fact, sent a kind of wave down the scissors,
carrying the information that the scissors have been closed.  But this wave
does not travel faster than the speed of light.  It will take at least one
year for the tips of the blades, at the far end of the scissors, to feel
any force whatsoever, and, ultimately, to come together to completely close
the scissors. 

	As a practical matter, this theoretical upper limit to the rigidity
of the metal in the scissors is *far* higher than the rigidity of any real
material, so it would, in practice, take much much longer to close a real
pair of metal scissors with blades as long as these. 

	One can analyze this problem microscopically as well.  The
electromagnetic force which binds the atoms of the scissors together
propagates at the speeds of light. So if you displace some set of atoms in
the scissor (such as the entire handles), the force will not propagate down
the scissor instantaneously, This means that a scissor this big *must*
cease to act as a rigid body. You can move parts of it without other parts
moving at the same time. It takes some finite time for the changing forces
on the scissor to propagate from atom to atom, letting the far tip of the
blades "know" that the scissors have been closed. 

	Caveat:

	The contact point where the two blades meet is not a physical
object.  So there is no fundamental reason why it could not move faster
than the speed of light, provided that you arrange your experiment correctly.
In fact it can be done with scissors provided that your scissors are short
enough and wide open to start, very different conditions than those spelled
out in the gedankenexperiment above.  In this case it will take you quite
a while to bring the blades together - more than enough time for light to
travel to the tips of the scissors.  When the blades finally come together,
if they have the right shape, the contact point can indeed move faster
than light.  

	Think about the simpler case of two rulers pinned together at an 
edge point at the ends.  Slam the two rulers together and the contact point
will move infinitely fast to the far end of the rulers at the instant
they touch.  So long as the rulers are short enough that contact does not
happen until the signal propagates to the far ends of the rulers, the 
rulers will indeed be straight when they meet.  Only if the rulers are
too long will they be bent like our very long scissors, above, when they
touch.  The contact point can move faster than the speed of light, but
the energy (or signal) of the closing force can not.

	An analogy, equivalent in terms of information content, is, say, a 
line of strobe lights.  You want to light them up one at a time, so that 
the `bright' spot travels faster than light.  To do so, you can send a 
_luminal_ signal down the line, telling each strobe light to wait a 
little while before flashing.  If you decrease the wait time with
each successive strobe light, the apparent bright spot will travel faster
than light, since the strobes on the end didn't wait as long after getting
the go-ahead, as did the ones at the beginning.  But the bright spot
can't pass the original signal, because then the strobe lights wouldn't
know to flash.


********************************************************************************

Item 25.  Can You See the Lorentz-Fitzgerald Contraction?           12-Oct-1995
          Or: Penrose-Terrell Rotation                              by Michael Weiss

People sometimes argue over whether the Lorentz-Fitzgerald contraction is
"real" or not.  That's a topic for another FAQ entry, but here's a short
answer: the contraction can be measured, but the measurement is
frame-dependent.  Whether that makes it "real" or not has more to do with your
choice of words than the physics.

Here we ask a subtly different question.  If you take a snapshot of a rapidly
moving object, will it *look* flattened when you develop the film?  What is the
difference between measuring and photographing?  Isn't seeing believing?  Not
always!  When you take a snapshot, you capture the light-rays that hit the
*film* at one instant (in the reference frame of the film).  These rays may
have left the *object* at different instants; if the object is moving with
respect to the film, then the photograph may give a distorted picture.
(Strictly speaking snapshots aren't instantaneous, but we're idealizing.)

Oddly enough, though Einstein published his famous relativity paper in
1905, and Fitzgerald proposed his contraction several years earlier,
no one seems to have asked this question until the late '50s.  Then
Roger Penrose and James Terrell independently discovered that the
object will *not* appear flattened [1,2].  People sometimes say that
the object appears rotated, so this effect is called the
Penrose-Terrell rotation.

Calling it a rotation can be a bit confusing though.  Rotating an object brings
its backside into view, but it's hard to see how a contraction could do that.
Among other things, this entry will try to explain in just what sense
the Penrose-Terrell effect is a "rotation".

It will clarify matters to imagine *two* snapshots of the same object, taken by
two cameras moving uniformly with respect to each other.  We'll call them *his*
camera and *her* camera.  The cameras pass through each other at the origin at
t=0, when they take their two snapshots.  Say that the object is at rest with
respect to his camera, and moving with respect to hers.  By analysing the
process of taking a snapshot, the meaning of "rotation" will become clearer.

How should we think of a snapshot?  Here's one way: consider a pinhole camera.
(Just one camera, for the moment.)  The pinhole is located at the origin, and
the film occupies a patch on a sphere surrounding the origin.  We'll ignore all
technical difficulties(!), and pretend that the camera takes full spherical
pictures: the film occupies the entire sphere.

We need more than just a pinhole and film, though: we also need a shutter.  At
t=0, the shutter snaps open for an instant to let the light-rays through the
pinhole; these spread out in all directions, and at t=1 (in the rest-frame of
the camera) paint a picture on the spherical film.

Let's call points in the snapshot *pixels*.  Each pixel gets its color due to
an event, namely a light-ray hitting the sphere at t=1.  Now let's consider his
& her cameras, as we said before.  We'll use t for his time, and t' for hers.
At t=t'=0, the two pinholes coincide at the origin, the two shutters snap
simultaneously, and the light rays spread out.  At t=1 for *his* camera, they
paint *his* pixels; at t'=1 for *her* camera, they paint *hers*.  So the
definition of a snapshot is frame-dependent.  But you already knew that.  (Pop
quiz: what shape does *he* think *her* film has?  Not spherical!)  (More
technical difficulties: the rays have to pass right through one film to hit the
other.)

So there's a one-one correspondence between pixels in the two snapshots.  Two
pixels correspond if they are painted by the same light-ray.  You can see now
that her snapshot is just a distortion of his (and vice versa).  You could take
his snapshot, scan it into a computer, run an algorithm to move the pixels
around, and print out hers.

So what does the pixel mapping look like?  Simple: if we put the usual
latitude/longitude grid on the spheres, chosen so that the relative motion is
along the north-south axis, then each pixel slides up towards the north pole
along a line of longitude.  (Or down towards the south pole, depending on
various choices I haven't specified.)  This should ring a bell if you know
about the aberration of light: if our snapshots portray the night-sky, then the
stars are white pixels, and aberration changes their apparent positions.

Now let's consider the object--- let's say a galaxy.  In passing from his
snapshot to hers, the image of the galaxy slides up the sphere, keeping the
same face to us.  In this sense, it has rotated.  Its apparent size will also
change, but not its shape (to a first approximation).

The mathematical details are beautiful, but best left to the textbooks [3,4].
Just to entice you if you have the background: if we regard the two spheres as
Riemann spheres, then the pixel mapping is given by a fractional linear
transformation.  Well-known facts from complex analysis now tell us two things.
First, circles go to circles under the pixel mapping, so a sphere will *always*
photograph as a sphere.  Second, shapes of objects are preserved in the
infinitesimally small limit.  (If you know about the double-covering of SL(2),
that also comes into play.  [3] is a good reference.)

References: [1] and [2] are the original articles.  [3] and [4] are textbook
treatments.  [5] has beautiful computer-generated pictures of the
Penrose-Terrell rotation.  The authors of [5] later made a video [6] of this
and other effects of "SR photography".

[1] Penrose, R.,"The Apparent Shape of a Relativistically Moving Sphere", 
    Proc. Camb. Phil. Soc., vol 55 Jul 1958.
[2] Terrell, J., "Invisibility of the Lorentz Contraction",
    Phys. Rev. vol 116 no. 4 pp. 1041-1045 (1959).    
[3] Penrose, R., and W. Rindler, "Spinors and Space-Time", vol I chapter 1.
[4] Marion, "Classical Dynamics", Section 10.5.
[5] Hsiung, Ping-Kang, Robert H. Thibadeau, and Robert H. P. Dunn,
    "Ray-Tracing Relativity", Pixel, vol 1 no. 1 (Jan/Feb 1990).
[6] Hsiung, Ping-Kang, and Robert H. Thibadeau, "Spacetime
    Visualizations," a video, Imaging Systems Lab, Robotics Institute,
    Carnegie Mellon University.



********************************************************************************

Item 26.  
Tachyons                                        updated: 22-MAR-1993 by SIC
--------                                        original by Scott I. Chase

                There was a young lady named Bright,
                Whose speed was far faster than light.
                She went out one day,
                In a relative way,
                And returned the previous night!

                        -Reginald Buller


	It is a well known fact that nothing can travel faster than the
speed of light. At best, a massless particle travels at the speed of light.
But is this really true?  In 1962, Bilaniuk, Deshpande, and Sudarshan, Am.
J. Phys. _30_, 718 (1962), said "no".  A very readable paper is Bilaniuk
and Sudarshan, Phys. Today _22_,43 (1969).  I give here a brief overview. 

	Draw a graph, with momentum (p) on the x-axis, and energy (E) on
the y-axis.  Then draw the "light cone", two lines with the equations E =
+/- p. This divides our 1+1 dimensional space-time into two regions.  Above
and below are the "timelike" quadrants, and to the left and right are the
"spacelike" quadrants. 

	Now the fundamental fact of relativity is that E^2 - p^2 = m^2. 
(Let's take c=1 for the rest of the discussion.)  For any non-zero value of 
m (mass), this is an hyperbola with branches in the timelike regions.  It 
passes through the point (p,E) = (0,m), where the particle is at rest.  Any 
particle with mass m is constrained to move on the upper branch of this 
hyperbola.  (Otherwise, it is "off-shell", a term you hear in association
with virtual particles - but that's another topic.) For massless particles, 
E^2 = p^2, and the particle moves on the light-cone. 

	These two cases are given the names tardyon (or bradyon in more
modern usage) and luxon, for "slow particle" and "light particle".  Tachyon
is the name given to the supposed "fast particle" which would move with v>c. 

	Now another familiar relativistic equation is E =
m*[1-(v/c)^2]^(-.5).  Tachyons (if they exist) have v > c.  This means that 
E is imaginary!  Well, what if we take the rest mass m, and take it to be
imaginary?  Then E is negative real, and E^2 - p^2 = m^2 < 0.  Or, p^2 -
E^2 = M^2, where M is real.  This is a hyperbola with branches in the
spacelike region of spacetime.  The energy and momentum of a tachyon must
satisfy this relation.

	You can now deduce many interesting properties of tachyons.  For
example, they accelerate (p goes up) if they lose energy (E goes down).
Futhermore, a zero-energy tachyon is "transcendent," or infinitely fast.
This has profound consequences.  For example, let's say that there were
electrically charged tachyons.  Since they would move faster than the speed 
of light in the vacuum, they should produce Cerenkov radiation. This would 
*lower* their energy, causing them to accelerate more!  In other words, 
charged tachyons would probably lead to a runaway reaction releasing an 
arbitrarily large amount of energy.  This suggests that coming up with a 
sensible theory of anything except free (noninteracting) tachyons is likely 
to be difficult.  Heuristically, the problem is that we can get spontaneous 
creation of tachyon-antitachyon pairs, then do a runaway reaction, making 
the vacuum unstable.  To treat this precisely requires quantum field theory, 
which gets complicated.  It is not easy to summarize results here.  However,
one reasonably modern reference is _Tachyons, Monopoles, and Related
Topics_, E. Recami, ed. (North-Holland, Amsterdam, 1978).

	However, tachyons are not entirely invisible.  You can imagine that
you might produce them in some exotic nuclear reaction.  If they are
charged, you could "see" them by detecting the Cerenkov light they produce
as they speed away faster and faster.  Such experiments have been done.  So
far, no tachyons have been found.  Even neutral tachyons can scatter off
normal matter with experimentally observable consequences.  Again, no such
tachyons have been found. 

	How about using tachyons to transmit information faster than the
speed of light, in violation of Special Relativity?  It's worth noting 
that when one considers the relativistic quantum mechanics of tachyons, the 
question of whether they "really" go faster than the speed of light becomes 
much more touchy!  In this framework, tachyons are *waves* that satisfy a
wave equation.  Let's treat free tachyons of spin zero, for simplicity.   
We'll set c = 1 to keep things less messy.  The wavefunction of a single 
such tachyon can be expected to satisfy the usual equation for spin-zero 
particles, the Klein-Gordon equation:

                (BOX + m^2)phi = 0

where BOX is the D'Alembertian, which in 3+1 dimensions is just

                BOX = (d/dt)^2 - (d/dx)^2 - (d/dy)^2 - (d/dz)^2.

The difference with tachyons is that m^2 is *negative*, and m is
imaginary.

To simplify the math a bit, let's work in 1+1 dimensions, with
coordinates x and t, so that

                BOX = (d/dt)^2 - (d/dx)^2

Everything we'll say generalizes to the real-world 3+1-dimensional case.
Now - regardless of m, any solution is a linear combination, or
superposition, of solutions of the form

                phi(t,x) = exp(-iEt + ipx)

where E^2 - p^2 = m^2.  When m^2 is negative there are two essentially
different cases.  Either |p| >= |E|, in which case E is real and
we get solutions that look like waves whose crests move along at the
rate |p|/|E| >= 1, i.e., no slower than the speed of light.  Or |p| <
|E|, in which case E is imaginary and we get solutions that look waves
that amplify exponentially as time passes!

We can decide as we please whether or not we want to consider the second
sort of solutions.   They seem weird, but then the whole business is
weird, after all.

1)	If we *do* permit the second sort of solution, we can solve the
Klein-Gordon equation with any reasonable initial data - that is, any
reasonable values of phi and its first time derivative at t = 0.  (For
the precise definition of "reasonable," consult your local
mathematician.)  This is typical of wave equations.  And, also typical
of wave equations, we can prove the following thing: If the solution phi
and its time derivative are zero outside the interval [-L,L] when t = 0,
they will be zero outside the interval [-L-|t|, L+|t|] at any time t.
In other words, localized disturbances do not spread with speed faster
than the speed of light!  This seems to go against our notion that
tachyons move faster than the speed of light, but it's a mathematical
fact, known as "unit propagation velocity".

2)	If we *don't* permit the second sort of solution, we can't solve the
Klein-Gordon equation for all reasonable initial data, but only for initial
data whose Fourier transforms vanish in the interval [-|m|,|m|].  By the
Paley-Wiener theorem this has an odd consequence: it becomes
impossible to solve the equation for initial data that vanish outside
some interval [-L,L]!  In other words, we can no longer "localize" our
tachyon in any bounded region in the first place, so it becomes
impossible to decide whether or not there is "unit propagation
velocity" in the precise sense of part 1).    Of course, the crests of
the waves exp(-iEt + ipx) move faster than the speed of light, but these
waves were never localized in the first place!

	The bottom line is that you can't use tachyons to send information 
faster than the speed of light from one place to another.  Doing so would 
require creating a message encoded some way in a localized tachyon field,
and sending it off at superluminal speed toward the intended receiver. But 
as we have seen you can't have it both ways - localized tachyon disturbances 
are subluminal and superluminal disturbances are nonlocal.

********************************************************************************
Item 27.

The Particle Zoo                                updated 4-JUL-1995 by MCW
----------------                                original by Matt Austern

	If you look in the Particle Data Book, you will find more than 150
particles listed there.  It isn't quite as bad as that, though... 

	The (observed) particles are divided into two major classes:
the material particles, and the gauge bosons.  We'll discuss the gauge
bosons further down.  The material particles in turn fall into three
categories: leptons, mesons, and baryons. Leptons are particles that
are like the electron: they have spin 1/2, and they do not undergo the
strong interaction.  There are three charged leptons, the electron,
muon, and tau, and three corresponding neutral leptons, or neutrinos.
(The muon and the tau are both short-lived.)

	Mesons and baryons both undergo strong interactions.  The
difference is that mesons have integral spin (0, 1,...), while baryons have
half-integral spin (1/2, 3/2,...).  The most familiar baryons are the
proton and the neutron; all others are short-lived.  The most familiar
meson is the pion; its lifetime is 26 nanoseconds, and all other mesons
decay even faster. 

	Most of those 150+ particles are mesons and baryons, or,
collectively, hadrons.  The situation was enormously simplified in the
1960s by the "quark model," which says that hadrons are made out of
spin-1/2 particles called quarks.  A meson, in this model, is made out
of a quark and an anti-quark, and a baryon is made out of three
quarks.  We don't see free quarks, but only hadrons; nevertheless, the
evidence for quarks is compelling. Quark masses are not very well
defined, since they are not free particles, but we can give estimates.
The masses below are in GeV; the first is current mass and the second
constituent mass (which includes some of the effects of the binding
energy):

      Generation:       1             2            3
      U-like:     u=.006/.311   c=1.50/1.65   t=91-200/91-200

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