Meeting Pearls 1/Meeting Pearls Vol 1 (1994).iso/installed_progs/text/faqs/physics-faq.part3

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Subject: Sci.physics Frequently Asked Questions - May 1994 - Part 3/4
Newsgroups: sci.physics,sci.physics.particle,alt.sci.physics.new-theories,news.answers,sci.answers,alt.answers
From: [e-mail] (SCOTT I CHASE)
Date: 30 Apr 1994 17:02 PST

Archive-name: physics-faq/part3
Last-modified: 26-APR-1994

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               FREQUENTLY ASKED QUESTIONS ON SCI.PHYSICS - Part 3/4
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Item 16.
                                                updated 9-DEC-1993 by SIC
                                                Original by Bill Johnson
How to Change Nuclear Decay Rates
---------------------------------

"I've had this idea for making radioactive nuclei decay faster/slower than
they normally do.  You do [this, that, and the other thing].  Will this work?"

Short Answer: Possibly, but probably not usefully.

Long Answer:

    "One of the paradigms of nuclear science since the very early days
of its study has been the general understanding that the half-life, or
decay constant, of a radioactive substance is independent of extranuclear
considerations."  (Emery, cited below.)  Like all paradigms, this one is
subject to some interpretation. Normal decay of radioactive stuff proceeds
via one of four mechanisms:

    * Emission of an alpha particle -- a helium-4 nucleus -- reducing
    the number of protons and neutrons present in the parent nucleus
    by two each;
    * "Beta decay," encompassing several related phenomena in which a
    neutron in the nucleus turns into a proton, or a proton turns into
    a neutron -- along with some other things including emission of
    a neutrino.  The "other things", as we shall see, are at the bottom
    of several questions involving perturbation of decay rates;
    * Emission of one or more gamma rays -- energetic photons -- that
    take a nucleus from an excited state to some other (typically
    ground) state; some of these photons may be replaced by
    "conversion electrons," of which more shortly; or
    *Spontaneous fission, in which a sufficiently heavy nucleus simply
    breaks in half.  Most of the discussion about alpha particles will
    also apply to spontaneous fission.

Gamma emission often occurs from the daughter of one of the other decay
modes.  We neglect *very* exotic processes like C-14 emission or double
beta decay in this analysis.

    "Beta decay" refers most often to a nucleus with a neutron excess,
which decays by converting a neutron into a proton:

         n ----> p + e- + anti-nu(e),

where n means neutron, p means proton, e- means electron, and anti-nu(e)
means an antineutrino of the electron type.  The type of beta decay which
involves destruction of a proton is not familiar to many people, so
deserves a little elaboration.  Either of two processes may occur when this
kind of decay happens:

        p ----> n + e+ + nu(e),

where e+ means positron and nu(e) means electron neutrino; or

        p + e- ----> n + nu(e),

where e- means a negatively charged electron, which is captured from the
neighborhood of the nucleus undergoing decay.  These processes are called
"positron emission" and "electron capture," respectively.  A given nucleus
which has too many protons for stability may undergo beta decay through
either, and typically both, of these reactions.

    "Conversion electrons" are produced by the process of "internal
conversion," whereby the photon that would normally be emitted in gamma
decay is *virtual* and its energy is absorbed by an atomic electron.  The
absorbed energy is sufficient to unbind the electron from the nucleus
(ignoring a few exceptional cases), and it is ejected from the atom as a
result.

    Now for the tie-in to decay rates.  Both the electron-capture and
internal conversion phenomena require an electron somewhere close to the
decaying nucleus.  In any normal atom, this requirement is satisfied in
spades: the innermost electrons are in states such that their probability
of being close to the nucleus is both large and insensitive to things in
the environment.  The decay rate depends on the electronic wavefunctions,
i.e, how much of their time the inner electrons spend very near the
nucleus -- but only very weakly. For most nuclides that decay by electron
capture or internal conversion, most of the time, the probability of
grabbing or converting an electron is also insensitive to the environment,
as the innermost electrons are the ones most likely to get grabbed/converted.

    However, there are exceptions, the most notable being the
the astrophysically important isotope beryllium-7.  Be-7 decays purely
by electron capture (positron emission being impossible because of
inadequate decay energy) with a half-life of somewhat over 50 days.  It has
been shown that differences in chemical environment result in half-life
variations of the order of 0.2%, and high pressures produce somewhat
similar changes. Other cases where known changes in decay rate occur are
Zr-89 and Sr-85, also electron capturers; Tc-99m ("m" implying an excited
state), which decays by both beta and gamma emission; and various other
"metastable" things that decay by gamma emission with internal conversion.
With all of these other cases the magnitude of the effect is less than is
typically the case with Be-7.

    What makes these cases special?  The answer is that one or another
of the usual starting assumptions -- insensitivity of electron wave
function near the nucleus to external forces, or availability of the
innermost electrons for capture/conversion -- are not completely valid.
Atomic beryllium only has 4 electrons to begin with, so that the "innermost
electrons" are also practically the *outermost* ones and therefore much
more sensitive to chemical effects than usual.  With most of the other
cases, there is so little energy available from the decay (as little as a
few electron volts; compare most radioactive decays, where hundreds or
thousands of *kilo*volts are released), courtesy of accidents of nuclear
structure, that the innermost electrons can't undergo internal conversion.
Remember that converting an electron requires dumping enough energy into it
to expel it from the atom (more or less); "enough energy," in context, is
typically some tens of keV, so they don't get converted at all in these
cases.  Conversion therefore works only on some of the outer electrons,
which again are more sensitive to the environment.

    A real anomaly is the beta emitter Re-187.  Its decay energy is
only about 2.6 keV, practically nothing by nuclear standards.  "That this
decay occurs at all is an example of the effects of the atomic environment
on nuclear decay: the bare nucleus Re-187 [i.e., stripped of all orbital
electrons -- MWJ] is stable against beta decay [but not to bound state
beta decay, in which the outgoing electron is captured by the daughter
nucleus into a tightly bound orbital -SIC] and it is the difference of
15 keV in the total electronic binding energy of osmium [to which it decays
-- MWJ] and rhenium ... which makes the decay possible" (Emery).  The
practical significance of this little peculiarity, of course, is low, as
Re-187 already has a half life of over 10^10 years.

    Alpha decay and spontaneous fission might also be affected by
changes in the electron density near the nucleus, for a different reason.
These processes occur as a result of penetration of the "Coulomb barrier"
that inhibits emission of charged particles from the nucleus, and their
rate is *very* sensitive to the height of the barrier.  Changes in the
electron density could, in principle, affect the barrier by some tiny
amount.  However, the magnitude of the effect is *very* small, according to
theoretical calculations; for a few alpha emitters, the change has been
estimated to be of the order of 1 part in 10^7 (!) or less, which would be
unmeasurable in view of the fact that the alpha emitters' half lives aren't
known to that degree of accuracy to begin with.

    All told, the existence of changes in radioactive decay rates due
to the environment of the decaying nuclei is on solid grounds both
experimentally and theoretically.  But the magnitude of the changes is
nothing to get very excited about.

Reference: The best review article on this subject is now 20 years old: G.
T. Emery, "Perturbation of Nuclear Decay Rates," Annual Review of Nuclear
Science vol. 22, p. 165 (1972).  Papers describing specific experiments are
cited in that article, which contains considerable arcane math but also
gives a reasonable qualitative "feel" for what is involved.

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

Item 17.
                        original by Blair P. Houghton
                        ([e-mail])

What is a Dippy Bird, and how is it used?
-----------------------------------------

The Anatomy and Habits of a Dippy Bird:

1.  The armature:  The body of the bird is a straight tube attached to two
bulbs, approximately the same size, one at either end.  The tube flows into
the upper bulb, like the neck of a funnel, and extends almost to the bottom
of the lower bulb, like the straw in a soda.

2.  The pivot:  At about the middle of the tube is clamped a transverse
bar, which allows the apparatus to pivot on a stand (the legs).  The bar is
bent very slightly concave dorsally, to unbalance the bird in the forward
direction (thus discouraging dips to the rear).  The ends of the pivot have
downward protrusions, which hit stops on the stand placed so that the bird
is free to rock when in a vertical position, but can not quite rotate
enough to be horizontal during a dip.

3.  The wick:  The upper bulb is coated in fuzzy material, and has extended
from it a beak, made of or covered in the same material.

4.  The tail.  The tail has no significant external features, except that
it should not be insulated (skin-oil deposited on the bird's glass parts
from handling will insulate it and can affect its operation).

5.  The guts:  The bird is partially filled with a somewhat carefully
measured amount of a fluid with suitable lack of viscosity and density and
a low latent heat of evaporation (small d(energy)/d(mass), ld).  For water,
ld is 2250 kJ/kg; for methylene chloride, ld is 406; for mercury, ld is a
wondrous 281; ethyl alcohol has an ld of 880, more than twice that of MC.
Boiling point is not important, here; evaporation and condensation take
place on the surface of a liquid at any temperature.

6.  The frills:  Any hats, eyes, feathers, or liquid coloring have been
added purely for entertainment value. (An anecdote:  as it stood pumping in
the Arizona sun on my kitchen windowsill for several days, the rich,
Kool-Aid red of my bird's motorwater faded to a pale peach.  I have since
retired him to the mantelpiece in the family room).

7.  Shreddin':  The bird is operated by getting the head wet, taking care
not to make it so wet that it drips down the tube.  (Water on the bottom
bulb will reverse the thermodynamic processes.)  The first cycle will
take somewhat longer than the following cycles.  If you can keep water
where the bird can dip it, the bird will dip for as long as the ambient
humidity remains favorable.


Come on, how does it really work?
---------------------------------

Short answer:  Thermodynamics plus Mechanics.

Medium answer (and essential clues):  Evaporative cooling on the outside;
pV=nRT, evaporation/condensation, and gravity on the inside.

Long answer:

    Initially the system is at equilibrium, with T equal in both
chambers and pV/n in each compensating for the fluid levels.  Evaporation
of water outside the head draws heat from inside it; the vapor inside
condenses, reducing pV/RT.  This imbalances the pressures, so the vapor in
the abdomen pushes down, which pushes fluid up the thorax, which reduces V
in the head.  Since p is decreasing in the abdomen, evaporation occurs,
increasing n, and drawing heat from outside the body.

    The rising fluid raises the CM above the pivot point; the hips are
slightly concave dorsally, so the bird dips forward.  Tabs on the legs and
the pivot maintain the angle at full dip, for drainage.  The amount of
fluid is set so that at full dip the lower end of the tube is exposed to
the vapor.  (The tube reaches almost to the bottom of the abdomen, like a
straw in a soda, but flows into the head like the neck of a funnel.)  A
bubble of vapor rises in the tube and fluid drains into the abdomen.

    The rising bubble transfers heat to the head and the falling fluid
releases gravitational potential energy as heat into the rising bubble and
the abdomen.  The CM drops below the pivot point and the bird bobs up.  The
system is thus reset; it's not quite at equilibrium, but is close enough
that the process can repeat this chain of events.

    The beak acts as a wick, if allowed to dip into a reservoir of
water, to keep the head wet, although it is not necessary for the bird to
drink on every dip.


Is that all there is to know about dippy birds?
-----------------------------------------------

    Of course not.  Research continues to unravel these unanswered
questions about the amazing dippy-bird:

1.  All of the energy gained by the rising fluid is returned to the system
when the fluid drops; where does this energy go, in what proportions, and
how does this affect the rate at which the bird operates?

2.  The heat that evaporates the water comes from both the surrounding air
and the inside of the head; but, in what proportion?

3.  Exactly what should the fluid be?  Methylene Chloride is an excellent
candidate, since it's listed in the documentation for recent birds sold by
Edmund Scientific Corp. (trade named Happy Drinking Bird), and because its
latent heat of evaporation (ld) is 406 kJ/kg, compared to 2250 kJ/kg for
water (a 5.5:1 ratio of condensed MC to evaporated water, if all
water-evaporating heat comes from inside the bird).  Ethanol, at 880 kJ/kG,
is only half as efficient.  Mercury would likewise be a good prospective
choice, having an ld of 281 kJ/kG (8:1!), but is expensive and dangerous,
and its density would require careful redesign and greater quality control
in the abdomen and pivot-stops to ensure proper operation at full dip; this
does, however, indicate that the apparatus could be made in miniature,
filled with mercury, and sold through a catalog-store such as The Sharper
Image as a wildly successful yuppie desk-toy (Consider the submission of
this FAQ entry to be prior art for patent purposes).

4.  Does ambient temperature have an effect on operation aside from the
increase in rate of evaporation of water? I.e., if the temperature and
humidity can be controlled independently such that the rate of evaporation
can be kept constant, what effect does such a change in ambient temperature
and humidity have on the operation of the bird? Is the response transient,
permanent, or composed of both?

Dippy Bird Tips:
----------------

    They have real trouble working at all in humid climates (like
around the U. of Md., where I owned my first one), but can drive you bats
in dry climates (aside from the constant hammering, it's hard to keep the
water up to a level where the bird can get at it...).  The evaporation of
water from the head depends on the diffusibility of water vapor into the
atmosphere; high partial pressures of water vapor in the atmosphere
translate to low rates of evaporation.

    If you handle your bird, clean the glass with alcohol or Windex
or Dawn or something; the oil from your hands has a high specific heat,
which damps the transfer of heat, and a low thermal conductivity, which
attenuates the transfer of heat. Once it's clean, grasp the bird only by
the legs or the tube, which are not thermodynamically significant, or
wear rubber gloves, just like a real EMT.

    The hat is there for show; the dippy bird operates okay with or
without it, even though it may reduce the area of evaporation slightly.
Ditto the feathers and the eyes.

Bibliography:
-------------

    Chemical data from Gieck, K., _Engineering Formulas_, 3d. Ed.,
McGraw-Hill, 1979, as translated by J. Walters, B. Sc.

    I've also heard that SciAm had an "Amateur Scientist" column on
this technology a few years ago.  Perhaps someone who understands how a
library works could look up the yr and vol...

    Kool-Aid is a trademark of some huge corporation that makes its
money a farthing at a time...

********************************************************************************
Item 18.

Some Frequently Asked Questions About Black Holes   updated 2-JUL-1993 by MM
-------------------------------------------------   original by Matt McIrvin

Contents:

1. What is a black hole, really?
2. What happens to you if you fall in?
3. Won't it take forever for you to fall in?  Won't it take forever
   for the black hole to even form?
4. Will you see the universe end?
5. What about Hawking radiation?  Won't the black hole evaporate
   before you get there?
6. How does the gravity get out of the black hole?
7. Where did you get that information?

1. What is a black hole, really?

    In 1916, when general relativity was new, Karl Schwarzschild worked
out a useful solution to the Einstein equation describing the evolution of
spacetime geometry.  This solution, a possible shape of spacetime, would
describe the effects of gravity *outside* a spherically symmetric,
uncharged, nonrotating object (and would serve approximately to describe
even slowly rotating objects like the Earth or Sun).  It worked in much the
same way that you can treat the Earth as a point mass for purposes of
Newtonian gravity if all you want to do is describe gravity *outside* the
Earth's surface.

    What such a solution really looks like is a "metric," which is a
kind of generalization of the Pythagorean formula that gives the length of
a line segment in the plane.  The metric is a formula that may be used to
obtain the "length" of a curve in spacetime.  In the case of a curve
corresponding to the motion of an object as time passes (a "timelike
worldline,") the "length" computed by the metric is actually the elapsed
time experienced by an object with that motion.  The actual formula depends
on the coordinates chosen in which to express things, but it may be
transformed into various coordinate systems without affecting anything
physical, like the spacetime curvature.  Schwarzschild expressed his metric
in terms of coordinates which, at large distances from the object,
resembled spherical coordinates with an extra coordinate t for time.
Another coordinate, called r, functioned as a radial coordinate at large
distances; out there it just gave the distance to the massive object.

    Now, at small radii, the solution began to act strangely.  There
was a "singularity" at the center, r=0, where the curvature of spacetime
was infinite.  Surrounding that was a region where the "radial" direction
of decreasing r was actually a direction in *time* rather than in space.
Anything in that region, including light, would be obligated to fall toward
the singularity, to be crushed as tidal forces diverged. This was separated
from the rest of the universe by a place where Schwarzschild's coordinates
blew up, though nothing was wrong with the curvature of spacetime there.
(This was called the Schwarzschild radius.  Later, other coordinate systems
were discovered in which the blow-up didn't happen; it was an artifact of
the coordinates, a little like the problem of defining the longitude of the
North Pole.  The physically important thing about the Schwarzschild radius
was not the coordinate problem, but the fact that within it the direction
into the hole became a direction in time.)

    Nobody really worried about this at the time, because there was no
known object that was dense enough for that inner region to actually be
outside it, so for all known cases, this odd part of the solution would not
apply.  Arthur Stanley Eddington considered the possibility of a dying star
collapsing to such a density,  but rejected it as aesthetically unpleasant
and proposed that some new physics must intervene.  In 1939, Oppenheimer
and Snyder finally took seriously the possibility that stars a few times
more massive than the sun might be doomed to collapse to such a state at
the end of their lives.

    Once the star gets smaller than the place where Schwarzschild's
coordinates fail (called the Schwarzschild radius for an uncharged,
nonrotating object, or the event horizon) there's no way it can avoid
collapsing further.  It has to collapse all the way to a singularity for
the same reason that you can't keep from moving into the future! Nothing
else that goes into that region afterward can avoid it either, at least in
this simple case.  The event horizon is a point of no return.

    In 1971 John Archibald Wheeler named such a thing a black hole,
since light could not escape from it.  Astronomers have many candidate
objects they think are probably black holes, on the basis of several kinds
of evidence (typically they are dark objects whose large mass can be
deduced from their gravitational effects on other objects, and which
sometimes emit X-rays, presumably from infalling matter).  But the
properties of black holes I'll talk about here are entirely theoretical.
They're based on general relativity, which is a theory that seems supported
by available evidence.

2. What happens to you if you fall in?

    Suppose that, possessing a proper spacecraft and a self-destructive
urge, I decide to go black-hole jumping and head for an uncharged,
nonrotating ("Schwarzschild") black hole.  In this and other kinds of hole,
I won't, before I fall in, be able to see anything within the event
horizon.  But there's nothing *locally* special about the event horizon;
when I get there it won't seem like a particularly unusual place, except
that I will see strange optical distortions of the sky around me from all
the bending of light that goes on.  But as soon as I fall through, I'm
doomed.  No bungee will help me, since bungees can't keep Sunday from
turning into Monday.  I have to hit the singularity eventually, and before
I get there there will be enormous tidal forces-- forces due to the
curvature of spacetime-- which will squash me and my spaceship in some
directions and stretch them in another until I look like a piece of
spaghetti.  At the singularity all of present physics is mute as to what
will happen, but I won't care.  I'll be dead.

    For ordinary black holes of a few solar masses, there are actually
large tidal forces well outside the event horizon, so I probably wouldn't
even make it into the hole alive and unstretched.  For a black hole of 8
solar masses, for instance, the value of r at which tides become fatal is
about 400 km, and the Schwarzschild radius is just 24 km.  But tidal
stresses are proportional to M/r^3.  Therefore the fatal r goes as the cube
root of the mass, whereas the Schwarzschild radius of the black hole is
proportional to the mass.  So for black holes larger than about 1000 solar
masses I could probably fall in alive, and for still larger ones I might
not even notice the tidal forces until I'm through the horizon and doomed.

3. Won't it take forever for you to fall in?  Won't it take forever
   for the black hole to even form?

    Not in any useful sense.  The time I experience before I hit the
event horizon, and even until I hit the singularity-- the "proper time"
calculated by using Schwarzschild's metric on my worldline -- is finite.
The same goes for the collapsing star; if I somehow stood on the surface of
the star as it became a black hole, I would experience the star's demise in
a finite time.

    On my worldline as I fall into the black hole, it turns out that
the Schwarzschild coordinate called t goes to infinity when I go through
the event horizon.  That doesn't correspond to anyone's proper time,
though; it's just a coordinate called t.  In fact, inside the event
horizon, t is actually a *spatial* direction, and the future corresponds
instead to decreasing r.  It's only outside the black hole that t even
points in a direction of increasing time.  In any case, this doesn't
indicate that I take forever to fall in, since the proper time involved is
actually finite.

    At large distances t *does* approach the proper time of someone who
is at rest with respect to the black hole.  But there isn't any
non-arbitrary sense in which you can call t at smaller r values "the proper
time of a distant observer," since in general relativity there is no
coordinate-independent way to say that two distant events are happening "at
the same time."  The proper time of any observer is only defined locally.

    A more physical sense in which it might be said that things take
forever to fall in is provided by looking at the paths of emerging light
rays.  The event horizon is what, in relativity parlance, is called a
"lightlike surface"; light rays can remain there.  For an ideal
Schwarzschild hole (which I am considering in this paragraph) the horizon
lasts forever, so the light can stay there without escaping.  (If you
wonder how this is reconciled with the fact that light has to travel at the
constant speed c-- well, the horizon *is* traveling at c! Relative speeds
in GR are also only unambiguously defined locally, and if you're at the
event horizon you are necessarily falling in; it comes at you at the speed
of light.)  Light beams aimed directly outward from just outside the
horizon don't escape to large distances until late values of t.  For
someone at a large distance from the black hole and approximately at rest
with respect to it, the coordinate t does correspond well to proper time.

    So if you, watching from a safe distance, attempt to witness my
fall into the hole, you'll see me fall more and more slowly as the light
delay increases.  You'll never see me actually *get to* the event horizon.
My watch, to you, will tick more and more slowly, but will never reach the
time that I see as I fall into the black hole.  Notice that this is really
an optical effect caused by the paths of the light rays.

    This is also true for the dying star itself.  If you attempt to
witness the black hole's formation, you'll see the star collapse more and
more slowly, never precisely reaching the Schwarzschild radius.

    Now, this led early on to an image of a black hole as a strange
sort of suspended-animation object, a "frozen star" with immobilized
falling debris and gedankenexperiment astronauts hanging above it in
eternally slowing precipitation.  This is, however, not what you'd see. The
reason is that as things get closer to the event horizon, they also get
*dimmer*.  Light from them is redshifted and dimmed, and if one considers
that light is actually made up of discrete photons, the time of escape of
*the last photon* is actually finite, and not very large.  So things would
wink out as they got close, including the dying star, and the name "black
hole" is justified.

    As an example, take the eight-solar-mass black hole I mentioned
before.  If you start timing from the moment the you see the object half a
Schwarzschild radius away from the event horizon, the light will dim
exponentially from that point on with a characteristic time of about 0.2
milliseconds, and the time of the last photon is about a hundredth of a
second later.  The times scale proportionally to the mass of the black
hole.  If I jump into a black hole, I don't remain visible for long.

    Also, if I jump in, I won't hit the surface of the "frozen star."
It goes through the event horizon at another point in spacetime from
where/when I do.

    (Some have pointed out that I really go through the event horizon a
little earlier than a naive calculation would imply.  The reason is that my
addition to the black hole increases its mass, and therefore moves the
event horizon out around me at finite Schwarzschild t coordinate.  This
really doesn't change the situation with regard to whether an external
observer sees me go through, since the event horizon is still lightlike;
light emitted at the event horizon or within it will never escape to large
distances, and light emitted just outside it will take a long time to get
to an observer, timed, say, from when the observer saw me pass the point
half a Schwarzschild radius outside the hole.)

    All this is not to imply that the black hole can't also be used for
temporal tricks much like the "twin paradox" mentioned elsewhere in this
FAQ.  Suppose that I don't fall into the black hole-- instead, I stop and
wait at a constant r value just outside the event horizon, burning
tremendous amounts of rocket fuel and somehow withstanding the huge
gravitational force that would result.  If I then return home, I'll have
aged less than you.  In this case, general relativity can say something
about the difference in proper time experienced by the two of us, because
our ages can be compared *locally* at the start and end of the journey.

4. Will you see the universe end?

    If an external observer sees me slow down asymptotically as I fall,
it might seem reasonable that I'd see the universe speed up
asymptotically-- that I'd see the universe end in a spectacular flash as I
went through the horizon.  This isn't the case, though.  What an external
observer sees depends on what light does after I emit it.  What I see,
however, depends on what light does before it gets to me.  And there's no
way that light from future events far away can get to me.  Faraway events
in the arbitrarily distant future never end up on my "past light-cone," the
surface made of light rays that get to me at a given time.

    That, at least, is the story for an uncharged, nonrotating black
hole.  For charged or rotating holes, the story is different.  Such holes
can contain, in the idealized solutions, "timelike wormholes" which serve
as gateways to otherwise disconnected regions-- effectively, different
universes.  Instead of hitting the singularity, I can go through the
wormhole.  But at the entrance to the wormhole, which acts as a kind of
inner event horizon, an infinite speed-up effect actually does occur.  If I
fall into the wormhole I see the entire history of the universe outside
play itself out to the end.  Even worse, as the picture speeds up the light
gets blueshifted and more energetic, so that as I pass into the wormhole an
"infinite blueshift" happens which fries me with hard radiation.  There is
apparently good reason to believe that the infinite blueshift would imperil
the wormhole itself, replacing it with a singularity no less pernicious
than the one I've managed to miss.  In any case it would render wormhole
travel an undertaking of questionable practicality.

5. What about Hawking radiation?  Won't the black hole evaporate
   before you get there?

    (First, a caveat: Not a lot is really understood about evaporating
black holes.  The following is largely deduced from information in Wald's
GR text, but what really happens-- especially when the black hole gets very
small-- is unclear.  So take the following with a grain of salt.)

Short answer:  No, it won't.  This demands some elaboration.

    From thermodynamic arguments Stephen Hawking realized that a black
hole should have a nonzero temperature, and ought therefore to emit
blackbody radiation.  He eventually figured out a quantum- mechanical
mechanism for this.  Suffice it to say that black holes should very, very
slowly lose mass through radiation, a loss which accelerates as the hole
gets smaller and eventually evaporates completely in a burst of radiation.
This happens in a finite time according to an outside observer.

    But I just said that an outside observer would *never* observe an
object actually entering the horizon!  If I jump in, will you see the black
hole evaporate out from under me, leaving me intact but marooned in the
very distant future from gravitational time dilation?

    You won't, and the reason is that the discussion above only applies
to a black hole that is not shrinking to nil from evaporation. Remember
that the apparent slowing of my fall is due to the paths of outgoing light
rays near the event horizon.  If the black hole *does* evaporate, the delay
in escaping light caused by proximity to the event horizon can only last as
long as the event horizon does!  Consider your external view of me as I
fall in.

    If the black hole is eternal, events happening to me (by my watch)
closer and closer to the time I fall through happen divergingly later
according to you (supposing that your vision is somehow not limited by the
discreteness of photons, or the redshift).

    If the black hole is mortal, you'll instead see those events happen
closer and closer to the time the black hole evaporates.  Extrapolating,
you would calculate my time of passage through the event horizon as the
exact moment the hole disappears!  (Of course, even if you could see me,
the image would be drowned out by all the radiation from the evaporating
hole.)  I won't experience that cataclysm myself, though; I'll be through
the horizon, leaving only my light behind. As far as I'm concerned, my
grisly fate is unaffected by the evaporation.

    All of this assumes you can see me at all, of course.  In practice
the time of the last photon would have long been past.  Besides, there's
the brilliant background of Hawking radiation to see through as the hole
shrinks to nothing.

    (Due to considerations I won't go into here, some physicists think
that the black hole won't disappear completely, that a remnant hole will be
left behind.  Current physics can't really decide the question, any more
than it can decide what really happens at the singularity. If someone ever
figures out quantum gravity, maybe that will provide an answer.)

6. How does the gravity get out of the black hole?

    Purely in terms of general relativity, there is no problem here.
The gravity doesn't have to get out of the black hole.  General relativity
is a local theory, which means that the field at a certain point in
spacetime is determined entirely by things going on at places that can
communicate with it at speeds less than or equal to c.  If a star collapses
into a black hole, the gravitational field outside the black hole may be
calculated entirely from the properties of the star and its external
gravitational field *before* it becomes a black hole. Just as the light
registering late stages in my fall takes longer and longer to get out to
you at a large distance, the gravitational consequences of events late in
the star's collapse take longer and longer to ripple out to the world at
large.  In this sense the black hole *is* a kind of "frozen star": the
gravitational field is a fossil field.  The same is true of the
electromagnetic field that a black hole may possess.

    Often this question is phrased in terms of gravitons, the
hypothetical quanta of spacetime distortion.  If things like gravity
correspond to the exchange of "particles" like gravitons, how can they get
out of the event horizon to do their job?

    First of all, it's important to realize that gravitons are not as
yet even part of a complete theory, let alone seen experimentally.  They
don't exist in general relativity, which is a non-quantum theory.  When
fields are described as mediated by particles, that's quantum theory, and
nobody has figured out how to construct a quantum theory of gravity.  Even
if such a theory is someday built, it may not involve "virtual particles"
in the same way other theories do.  In quantum electrodynamics, the static
forces between particles are described as resulting from the exchange of
"virtual photons," but the virtual photons only appear when one expresses
QED in terms of a quantum- mechanical approximation method called
perturbation theory.  It currently looks like this kind of perturbation
theory doesn't work properly when applied to quantum gravity.  So although
quantum gravity may well involve "real gravitons" (quantized gravitational
waves), it may well not involve "virtual gravitons" as carriers of static
gravitational forces.

    Nevertheless, the question in this form is still worth asking,
because black holes *can* have static electric fields, and we know that
these may be described in terms of virtual photons.  So how do the virtual
photons get out of the event horizon?  The answer is that virtual particles
aren't confined to the interiors of light cones: they can go faster than
light!  Consequently the event horizon, which is really just a surface that
moves at the speed of light, presents no barrier.

    I couldn't use these virtual photons after falling into the hole to
communicate with you outside the hole; nor could I escape from the hole by
somehow turning myself into virtual particles.  The reason is that virtual
particles don't carry any *information* outside the light cone.  That is a
tricky subject for another (future?) FAQ entry.  Suffice it to say that the
reasons virtual particles don't provide an escape hatch for a black hole
are the same as the reasons they can't be used for faster-than-light travel
or communication.

7. Where did you get that information?

    The numbers concerning fatal radii, dimming, and the time of the
last photon came from Misner, Thorne, and Wheeler's _Gravitation_ (San
Francisco: W. H. Freeman & Co., 1973), pp. 860-862 and 872-873. Chapters 32
and 33 (IMHO, the best part of the book) contain nice descriptions of some
of the phenomena I've described.

    Information about evaporation and wormholes came from Robert Wald's
_General Relativity_ (Chicago: University of Chicago Press, 1984). The
famous conformal diagram of an evaporating hole on page 413 has resolved
several arguments on sci.physics (though its veracity is in question).

    Steven Weinberg's _Gravitation and Cosmology_ (New York: John Wiley
and Sons, 1972) provided me with the historical dates.  It discusses some
properties of the Schwarzschild solution in chapter 8 and describes
gravitational collapse in chapter 11.

********************************************************************************
Item 19.

Below Absolute Zero - What Does Negative Temperature Mean?   updated 24-MAR-1993
----------------------------------------------------------   by Scott I. Chase

Questions:  What is negative temperature?  Can you really make a system
which has a temperature below absolute zero?  Can you even give any useful
meaning to the expression 'negative absolute temperature'?

Answer:  Absolutely. :-)

    Under certain conditions, a closed system *can* be described by a
negative temperature, and, surprisingly, be *hotter* than the same system
at any positive temperature.  This article describes how it all works.

Step I: What is "Temperature"?
------------------------------

    To get things started, we need a clear definition of "temperature."
Our intuitive notion is that two systems in thermal contact should exchange
no heat, on average, if and only if they are at the same temperature.  Let's
call the two systems S1 and S2. The combined system, treating S1 and S2
together, can be S3.  The important question, consideration of which
will lead us to a useful quantitative definition of temperature, is "How will
the energy of S3 be distributed between S1 and S2?"  I will briefly explain
this below, but I recommend that you read K&K, referenced below, for a
careful, simple, and thorough explanation of this important and fundamental
result.

    With a total energy E, S has many possible internal states
(microstates).  The atoms of S3 can share the total energy in many ways.
Let's say there are N different states.  Each state corresponds to a
particular division of the total energy in the two subsystems S1 and S2.
Many microstates can correspond to the same division, E1 in S1 and E2 in
S2. A simple counting argument tells you that only one particular division
of the energy, will occur with any significant probability.  It's the one
with the overwhelmingly largest number of microstates for the total system
S3. That number, N(E1,E2) is just the product of the number of states
allowed in each subsystem, N(E1,E2) = N1(E1)*N2(E2), and, since E1 + E2 =
E, N(E1,E2) reaches a maximum when N1*N2 is stationary with respect to
variations of E1 and E2 subject to the total energy constraint.

    For convenience, physicists prefer to frame the question in terms
of the logarithm of the number of microstates N, and call this the entropy,
S. You can easily see from the above analysis that two systems are in
equilibrium with one another when (dS/dE)_1 = (dS/dE)_2, i.e., the rate of
change of entropy, S, per unit change in energy, E, must be the same for
both systems.  Otherwise, energy will tend to flow from one subsystem to
another as S3 bounces randomly from one microstate to another, the total
energy E3 being constant, as the combined system moves towards a state of
maximal total entropy.  We define the temperature, T, by 1/T = dS/dE, so
that the equilibrium condition becomes the very simple T_1 = T_2.

    This statistical mechanical definition of temperature does in fact
correspond to your intuitive notion of temperature for most systems. So
long as dS/dE is always positive, T is always positive.  For common
situations, like a collection of free particles, or particles in a harmonic
oscillator potential, adding energy always increases the number of
available microstates, increasingly faster with increasing total energy. So
temperature increases with increasing energy, from zero, asymptotically
approaching positive infinity as the energy increases.

Step II: What is "Negative Temperature"?
----------------------------------------

    Not all systems have the property that the entropy increases
monotonically with energy.  In some cases, as energy is added to the system,
the number of available microstates, or configurations, actually decreases
for some range of energies.  For example, imagine an ideal "spin-system", a
set of N atoms with spin 1/2 one a one-dimensional wire.  The atoms are not
free to move from their positions on the wire.  The only degree of freedom
allowed to them is spin-flip:  the spin of a given atom can point up or
down.  The total energy of the system, in a magnetic field of strength B,
pointing down, is (N+ - N-)*uB, where u is the magnetic moment of each atom
and N+ and N- are the number of atoms with spin up and down respectively.
Notice that with this definition, E is zero when half of the spins are
up and half are down.  It is negative when the majority are down and
positive when the majority are up.

    The lowest possible energy state, all the spins will point down,
gives the system a total energy of -NuB, and temperature of absolute zero.
There is only one configuration of the system at this energy, i.e., all the
spins must point down.  The entropy is the log of the number of
microstates, so in this case is log(1) = 0.  If we now add a quantum of
energy, size uB, to the system, one spin is allowed to flip up.  There are
N possibilities, so the entropy is log(N).  If we add another quantum of
energy, there are a total of N(N-1)/2 allowable configurations with two
spins up.  The entropy is increasing quickly, and the temperature is rising
as well.

    However, for this system, the entropy does not go on increasing
forever.  There is a maximum energy, +NuB, with all spins up.  At this
maximal energy, there is again only one microstate, and the entropy is
again zero.  If we remove one quantum of energy from the system, we allow
one spin down.  At this energy there are N available microstates.  The
entropy goes on increasing as the energy is lowered.  In fact the maximal
entropy occurs for total energy zero, i.e., half of the spins up, half
down.

    So we have created a system where, as we add more and more energy,
temperature starts off positive, approaches positive infinity as maximum
entropy is approached, with half of all spins up.  After that, the
temperature becomes negative infinite, coming down in magnitude toward
zero, but always negative, as the energy increases toward maximum. When the
system has negative temperature, it is *hotter* than when it is has
positive system.  If you take two copies of the system, one with positive
and one with negative temperature, and put them in thermal contact, heat
will flow from the negative-temperature system into the positive-temperature
system.

Step III:  What Does This Have to Do With the Real World?
---------------------------------------------------------

    Can this system ever by realized in the real world, or is it just a
fantastic invention of sinister theoretical condensed matter physicists?
Atoms always have other degrees of freedom in addition to spin, usually
making the total energy of the system unbounded upward due to the
translational degrees of freedom that the atom has.  Thus, only certain
degrees of freedom of a particle can have negative temperature.  It makes
sense to define the "spin-temperature" of a collection of atoms, so long as
one condition is met:  the coupling between the atomic spins and the other
degrees of freedom is sufficiently weak, and the coupling between atomic
spins sufficiently strong, that the timescale for energy to flow from the
spins into other degrees of freedom is very large compared to the timescale
for thermalization of the spins among themselves.  Then it makes sense to
talk about the temperature of the spins separately from the temperature of
the atoms as a whole. This condition can easily be met for the case of
nuclear spins in a strong external magnetic field.

    Nuclear and electron spin systems can be promoted to negative
temperatures by suitable radio frequency techniques.   Various experiments
in the calorimetry of negative temperatures, as well as applications of
negative temperature systems as RF amplifiers, etc., can be found in the
articles listed below, and the references therein.

References:

        Kittel and Kroemer,_Thermal Physics_, appendix E.
        N.F. Ramsey, "Thermodynamics and statistical mechanics at negative
        absolute temperature,"  Phys. Rev. _103_, 20 (1956).
        M.J. Klein,"Negative Absolute Temperature," Phys. Rev. _104_, 589 (1956).

********************************************************************************
Item 20.

Which Way Will my Bathtub Drain?                updated 16-MAR-1993 by SIC
--------------------------------                original by Matthew R. Feinstein

Question: Does my bathtub drain differently depending on whether I live
in the northern or southern hemisphere?

Answer: No.  There is a real effect, but it is far too small to be relevant
when you pull the plug in your bathtub.

    Because the earth rotates, a fluid that flows along the earth's
surface feels a "Coriolis" acceleration perpendicular to its velocity.
In the northern hemisphere low pressure storm systems spin counterclockwise.
In the southern hemisphere, they spin clockwise because the direction
of the Coriolis acceleration is reversed.  This effect leads to the
speculation that the bathtub vortex that you see when you pull the plug
from the drain spins one way in the north and the other way in the south.

    But this acceleration is VERY weak for bathtub-scale fluid
motions.  The order of magnitude of the Coriolis acceleration can be
estimated from size of the "Rossby number" (see below).  The effect of the
Coriolis acceleration on your bathtub vortex is SMALL.  To detect its
effect on your bathtub, you would have to get out and wait until the motion
in the water is far less than one rotation per day.  This would require
removing thermal currents, vibration, and any other sources of noise.  Under
such conditions, never occurring in the typical home, you WOULD see an
effect.  To see what trouble it takes to actually see the effect, see the
reference below.  Experiments have been done in both the northern and
southern hemispheres to verify that under carefully controlled conditions,
bathtubs drain in opposite directions due to the Coriolis acceleration from
the Earth's rotation.

    Coriolis accelerations are significant when the Rossby number is
SMALL.     So, suppose we want a Rossby number of 0.1 and a bathtub-vortex
length scale of 0.1 meter.  Since the earth's rotation rate is about
10^(-4)/second, the fluid velocity should be less than or equal to
2*10^(-6) meters/second.  This is a very small velocity.  How small is it?
Well, we can take the analysis a step further and calculate another, more
famous dimensionless parameter, the Reynolds number.

    The Reynolds number is = L*U*density/viscosity

    Assuming that physicists bathe in hot water the viscosity will be
about 0.005 poise and the density will be about 1.0, so the Reynolds Number
is about 4*10^(-2).

    Now, life at low Reynolds numbers is different from life at high
Reynolds numbers.  In particular, at low Reynolds numbers, fluid physics is
dominated by friction and diffusion, rather than by inertia: the time it
would take for a particle of fluid to move a significant distance due to an
acceleration is greater than the time it takes for the particle to break up
due to diffusion.

    The same effect has been accused of responsibility for the
direction water circulates when you flush a toilet.  This is surely
nonsense.  In this case, the water rotates in the direction which the pipe
points which carries the water from the tank to the bowl.

Reference: Trefethen, L.M. et al, Nature 207 1084-5 (1965).

********************************************************************************
Item 21.                                         original by Scott I. Chase

Does Antimatter Fall Up or Down?
--------------------------------

This question has never been subject to a successful direct experiment.
In other words, nobody has ever directly measured the gravititational
acceleration of antimatter.  So the bottom line is that we don't know yet.
However, there is a lot more to say than just that, with regard to both
theory and experiment.  Here is a summary of the current state of affairs.

(1) Is is even theoretically possible for antimatter to fall up?

Answer: According to GR, antimatter falls down.

If you believe that General Relativity is the exact true theory of
gravity, then there is only one possible conclusion - by the equivalence
principle, antiparticles must fall down with the same acceleration as
normal matter.

On the other hand: there are other models of gravity which are not ruled out
by direct experiment which are distinct from GR in that antiparticles can
fall down at different rates than normal matter, or even fall up, due to
additional forces which couple to the mass of the particle in ways which are
different than GR.  Some people don't like to call these new couplings
'gravity.'  They call them, generically, the 'fifth force,' defining gravity
to be only the GR part of the force.  But this is mostly a semantic
distinction.  The bottom line is that antiparticles won't fall like normal
particles if one of these models is correct.

There are also a variety of arguments, based upon different aspects of
physics, against the possibility of antigravity.  These include constraints
imposed by conservation of energy (the "Morrison argument"), the detectable
effects of virtual antiparticles (the "Schiff argument"), and the absense
of gravitational effect in kaon regeneration experiments.  Each of these
does in fact rule out *some* models of antigravity.  But none of them
absolutely excludes all possible models of antigravity.  See the reference
below for all the details on these issues.

(2) Haven't people done experiments to study this question?

There are no valid *direct* experimental tests of whether antiparticles
fall up or down.  There was one well-known experiment by Fairbank at
Stanford in which he tried to measure the fall of positrons.  He found that
they fell normally, but later analyses of his experiment revealed that
he had not accounted for all the sources of stray electromagnetic fields.
Because gravity is so much weaker than EM, this is a difficult experimental
problem.  A modern assessment of the Fairbank experiment is that it was
inconclusive.

In order to reduce the effect of gravity, it would be nice to repeat the
Fairbank experiment using objects with the same magnitude of electric
charge as positrons, but with much more mass, to increase the relative
effect of gravity on the motion of the particle.  Antiprotons are 1836
times more massive than positrons, so give you three orders of magnitude
more sensitivity.  Unfortunately, making many slow antiprotons which you
can watch fall is very difficult.  An experiment is under development
at CERN right now to do just that, and within the next couple of years
the results should be known.

Most people expect that antiprotons *will* fall.  But it is important
to keep an open mind - we have never directly observed the effect of
gravity on antiparticles.  This experiment, if successful, will definitely
be "one for the textbooks."

Reference: Nieto and Goldman, "The Arguments Against 'Antigravity' and
the Gravitational Acceleration of Antimatter,"  Physics Reports, v.205,
No. 5, p.221.

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END OF FAQ PART 3/4