Subject: Re: Misused Units
From: Kent Budge <"kgbudge,"@,sandia,.gov>
Date: Feb 26 1997
Newsgroups: rec.arts.sf.science,sci.astro,sci.physics,alt.folklore.science,
	rec.arts.sf.tv

Mikko Juhani Vilenius wrote:
...
> I saw a document for quite a while ago about this.
> I believe it had something to do with the spin of electrons.
> 
> There was some material (hydrogen?) that was frozen very near to 0K
> and then put into powerful magneticfield, so that something happened
> to the spin of the electrons, which still dropped the energy level making
> the material "colder".
> 
> The scientists said that infact the energy level was increased radically,
> thus making the material "very hot".. This is where I drop off. I'm not
> too familiar with quantum mechanics and this was about four years ago.
> 
> Could somebody clarify this? Or did I get it totally wrong the first time?
...

What you are saying mostly makes sense, but *could* use some clarifying.
I'll give it a shot (takes deep breath and rubs sleepiness out of his
eyes):

The hydrogen nucleus (which is a single proton) has a quantum mechanical 
spin of 1/2. This gives it magnetic "north" and "south" poles, like 
a bar magnet.  Unlike a bar magnet, the proton can only have two
orientations with respect to a magnetic field; it either has its north
pole pointing towards the north direction of the field, or towards the
south.  Orientations in between simply don't exist, because the
orientation is "quantized" (that's where quantum mechanics gets its
name.)  Weird, but experimentally verified. (Hope I haven't lost you 
already *OR* insulted your intelligence -- I have no idea how extensive
your knowledge is.) 

When the north pole of the proton is pointed in the north direction, the
energy of the proton is greater than when the north pole of the proton
is pointed in the south direction.  (Like poles repel each other, so
it takes energy to force the north pole of the proton to point north.)
The energy difference depends on the strength of the magnetic field,
but it is quite small for ordinary field strengths.

Now let's look at a frozen hydrogen crystal.  The electrons are 
essentially all frozen into their lowest energy state, so we'll just
consider the protons.  Turn on the magnetic field.  Each proton will be
either aligned with or against the field.  If most are aligned with the
field, the total energy is high.  If most are aligned against the field,
the total energy is low.  

Now let's talk about temperature and statistical mechanics.  (very DEEP
breath.)  You can't possibly keep track of the sagans of protons in 
even a very tiny hydrogen crystal.  You have to examine the problem
statistically, which is where "statistical mechanics" comes in.  The
fundamental principle of statistical mechanics is:

	All accessible states of a system are equally probable.

In other words, for a given total energy within a system, all states of 
the system that have that same total energy are equally probable.  
This principle is based on the assumption that the system is constantly
moving between states more or less at random, so a snapshot at any
given moment is as likely to catch it in one state as another.

Suppose we have just enough energy in the hydrogen crystal to allow
half the protons to be aligned with the magnetic field.  There are
sagans of protons involved.  Let's suppose the crystal is very small,
so that there are just 2e20 protons involved.  (That's less than a
milligram of hydrogen.)  Think of all the ways you can divide up 
2e20 objects into two groups of 1e20 objects.  (Note to gurus:  yes,
I am aware that I am neglecting certain technical issues regarding the
distinguishability of like particles.  It doesn't matter for this
discussion.)  There are about (2e20)!/(1e20!)/(1e20!) = 1e(1.386e20) 
ways to do it.  That last quantity is meant to represent 1 followed
by 1.386e20 zeros.  Carl Sagan would be proud.  In other words, there
are 1e(1.386e20) states accessible to the crystal, each equally 
probable.

Now consider what happens if there is enough energy for all the protons
to be aligned with the field.  There's only one way to align all the
protons with the field; it's the state in which all the protons are
aligned with the field. ;-)  So there is exactly one accessible state,
with probability 1.000...  

The same is true when you have the minimum possible energy, forcing
all the protons to be aligned against the field.  There's only one
such state.

If you plot the number of accessible states versus the energy, you
get a function that is strongly peaked at the energy necessary to
align half the protons with the field.  (I mean *STRONGLY* peaked.)

The "entropy" of this system, incidentally, is proportional to the
logarithm of the number of accessible states.  So the entropy is highest
for the half-aligned system, and is zero for either the fully-aligned
or fully-unaligned state.

Suppose this system comes into contact with another system, so that
energy can flow between them.  Suppose it's another hydrogen crystal.
Suppose one is fully unaligned and the other is half aligned.  If no
energy flows, the one crystal has one accessible state and the other
has 1e(1.386e20) accessible states.  The number of accessible states 
for the combination of the two crystals is the product of these two
numbers, or 1e(1.386e20).  If, on the other hand, energy flows until the
two crystals each have an equal energy -- enough to align 25% of
the protons -- each crystal has (2e20)!/(5e19)!/(1.5e20)! = 1e(1.125e20)
accessible states.  The number of accessible states for the combination
of the two crystals is then 1e(2.249e20), way more than for the case
where no energy flows.  So its *extremely* probable that energy will
flow from the half-aligned crystal to the fully unaligned crystal until
the energy is equally shared between them.

Let U be the energy in a crystal.  Let S(U) be the log of the number
of accessible states of the crystal for an energy U (the "entropy.")
Then the log of the total number of accessible states for the 
combination of two crystals is S(U1) + S(U2).  (Note to gurus:  This
is why entropy is the *log* of the number of accessible states -- it
makes it an extensive quantity.)  Suppose we increase U2 at the expense 
of U1 (that is, energy flows to crystal 2 from crystal 1.)  If this
increases S(U2) more than it diminishes S(U1), the total number of
accessible states increases.  So such an energy exchange is likely to
take place.  If S(U2) is increased less than S(U1) is diminished, 
the total number of accessible states is decreased, and the energy 
exchange is unlikely to take place.  

We therefore define the temperature of the system to be

	1/T = dS/dU

You can see that a crystal with a higher temperature will tend to give
its energy to a crystal with a lower temperature, because the higher
temperature means a smaller reduction in entropy when energy is given
up, and entropy is the measure of the number of states accessible to
the system.

For most systems, dS/dU is always positive.  But for our hydrogen 
crystal, dS/dU is negative if the crystal has enough energy to align
more than half the protons.  So negative temperatures actually 
correspond to more energy than do positive temperatures.  

If you put a negative temperature crystal in contact with a positive
temperature crystal, the entropy of both crystals increases as energy
flows from the negative temperature crystal to the positive temperature
crystal.  So, odd as it seems, energy flows from negative temperature
systems to positive temperature systems.  The negative temperature
systems are "hotter."  Stranger still, the negative temperature system
becomes more negative in temperature as energy flows, and the rate of
change in temperature accelerates as more energy flows until the system
passes from infinitely negative temperature to infinitely positive
temperature in the blink of an eye.  This is purely an artifact of the 
weird reciprocal definition of temperature; the *energies* involved 
certainly aren't infinite.

I hope this helps, though I fear it may be more than anyone can digest
in a single reading.

-- 

-Kent

(usual disclaimer)
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