Item 15.

Below Absolute Zero - What Does Negative Temperature Mean?   updated

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

        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

        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

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.


        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

Item 16.

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

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 17.

Why do Mirrors Reverse Left and Right?          updated 16-MAR-1993 by SIC

        The simple answer is that they don't.  Look in a mirror and wave
your right hand.  On which side of the mirror is the hand that waved?  The
right side, of course.

        Mirrors DO reverse In/Out.  Imaging holding an arrow in your hand.
If you point it up, it will point up in the mirror.  If you point it to the
left, it will point to the left in the mirror.  But if you point it toward
the mirror, it will point right back at you.  In and Out are reversed.

        If you take a three-dimensional, rectangular, coordinate system,
(X,Y,Z), and point the Z axis such that the vector equation X x Y = Z is
satisfied, then the coordinate system is said to be right-handed.  Imagine
Z pointing toward the mirror.  X and Y are unchanged (remember the arrows?)
but Z will point back at you.  In the mirror, X x Y = - Z.  The image
contains a left-handed coordinate system.

        This has an important effect, familiar mostly to chemists and
physicists. It changes the chirality, or handedness of objects viewed in
the mirror. Your left hand looks like a right hand, while your right hand
looks like a left hand.  Molecules often come in pairs called
stereoisomers, which differ not in the sequence or number of atoms, but
only in that one is the mirror image of the other, so that no rotation or
stretching can turn one into the other.  Your hands make a good laboratory
for this effect.  They are distinct, even though they both have the same
components connected in the same way. They are a stereo pair, identical
except for "handedness".

        People sometimes think that mirrors *do* reverse left/right, and
that the effect is due to the fact that our eyes are aligned horizontally
on our faces.  This can be easily shown to be untrue by looking in any
mirror with one eye closed!

Reference:  _The Left Hand of the Neutrino_, by Isaac Asimov, contains
a very readable discussion of handedness and mirrors in physics.

Item 18.

What is the Mass of a Photon?                   updated 24-JUL-1992 by SIC
                                                original by Matt Austern

Or, "Does the mass of an object depend on its velocity?"

        This question usually comes up in the context of wondering whether
photons are really "massless," since, after all, they have nonzero energy.
The problem is simply that people are using two different definitions of
mass.  The overwhelming consensus among physicists today is to say that
photons are massless.  However, it is possible to assign a "relativistic
mass" to a photon which depends upon its wavelength.  This is based upon
an old usage of the word "mass" which, though not strictly wrong, is not
used much today.

        The old definition of mass, called "relativistic mass," assigns
a mass to a particle proportional to its total energy E, and involved
the speed of light, c, in the proportionality constant:

                m = E / c^2.                                        (1)

This definition gives every object a velocity-dependent mass.

        The modern definition assigns every object just one mass, an
invariant quantity that does not depend on velocity.  This is given by

                m = E_0 / c^2,                                      (2)

where E_0 is the total energy of that object at rest.

        The first definition is often used in popularizations, and in some
elementary textbooks.  It was once used by practicing physicists, but for
the last few decades, the vast majority of physicists have instead used the
second definition.  Sometimes people will use the phrase "rest mass," or
"invariant mass," but this is just for emphasis: mass is mass.  The
"relativistic mass" is never used at all.  (If you see "relativistic mass"
in your first-year physics textbook, complain! There is no reason for books
to teach obsolete terminology.)

        Note, by the way, that using the standard definition of mass, the
one given by Eq. (2), the equation "E = m c^2" is *not* correct.  Using the
standard definition, the relation between the mass and energy of an object
can be written as

                E   = m c^2 / sqrt(1 -v^2/c^2),                     (3)
or as

                E^2 = m^2 c^4  +  p^2 c^2,                          (4)

where v is the object's velocity, and p is its momentum.

        In one sense, any definition is just a matter of convention.  In
practice, though, physicists now use this definition because it is much
more convenient.  The "relativistic mass" of an object is really just the
same as its energy, and there isn't any reason to have another word for
energy: "energy" is a perfectly good word.  The mass of an object, though,
is a fundamental and invariant property, and one for which we do need a

        The "relativistic mass" is also sometimes confusing because it
mistakenly leads people to think that they can just use it in the Newtonian
                F = m a                                             (5)
                F = G m1 m2 / r^2.                                  (6)

In fact, though, there is no definition of mass for which these
equations are true relativistically: they must be generalized.  The
generalizations are more straightforward using the standard definition
of mass than using "relativistic mass."

        Oh, and back to photons: people sometimes wonder whether it makes
sense to talk about the "rest mass" of a particle that can never be at
rest.  The answer, again, is that "rest mass" is really a misnomer, and it
is not necessary for a particle to be at rest for the concept of mass to
make sense.  Technically, it is the invariant length of the particle's
four-momentum.  (You can see this from Eq. (4).)  For all photons this is
zero. On the other hand, the "relativistic mass" of photons is frequency
dependent. UV photons are more energetic than visible photons, and so are
more "massive" in this sense, a statement which obscures more than it

        Reference: Lev Okun wrote a nice article on this subject in the
June 1989 issue of Physics Today, which includes a historical discussion
of the concept of mass in relativistic physics.