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Re: [Phys-L] heat content



In high school terms, I would say that absolute zero is the temperature which represents (or some related word) the lowest possible energy condition for an object. For atoms, that means that at least the electrons are still moving. I don't know what absolute zero for a neutron star would mean, but they are fermions, so that has to provide a clue.

-----Original Message-----
From: Phys-l [mailto:phys-l-bounces@phys-l.org] On Behalf Of Anthony
Lapinski
Sent: Tuesday, February 18, 2014 9:30 AM
To: Phys-L@Phys-L.org
Subject: Re: [Phys-L] heat content

This is all very interesting but way beyond my high school classes that I teach
(any my physics knowledge!).

What do I tell kids about motion and absolute zero? What's appropriate?

They know that temperature is a measure of the average KE of the
molecules in an object. To them, absolute zero would imply no motion.



Phys-L@Phys-L.org writes:
Regarding JD's response:

The fact that a hydrogen atom has nonzero size is most easily
explained in terms of the zero-point motion of the electron.

On 02/17/2014 08:34 PM, David Bowman replied:

I think a fairly compelling argument can be made for the contrary
answer.

Let's just say I find the counterargument to be not compelling.
It looks to me like a bunch of word games, not physics.

So requiring that something actually *move* in order for motion to
exist is not compelling to you. To each their own. I agree the
disagreement is solely about words. I'm certain we both agree on the
physics.

But I believe which answer is actually correct tends to hinge on the
technical details of just what one considers to be the definition of
the notion of 'motion'.

Well, sure, but the actual correct answer is known, the technical
details are known, and the definition of 'motion' is known. These
things are not up for grabs.

True. However, an accepted definition of motion is "the action or
process of moving or being moved". Zero point motion has nothing
actually moving. Why is it then called motion? In a quantum ground
state (for a time-independent Hamiltonian) nothing changes its
position, its orientation, its shape, or any other property as a function of
time.
Nothing is moving.

There are about ten lines of reasoning that all lead to the same
answer. Since yesterday's explanation evidently didn't suffice, let
me come at it from another angle.

Consider the a bunch of non-interacting particles in the gas phase.
The concept of pressure is pretty well understood.
It has to do with the motion of the particles. You can put scare
quotes around 'motion' but it doesn't scare me;

I didn't mean to scare anyone. I was merely identifying or pointing
out the word whose definition is not satisfied by the phrase 'zero
point motion'.

motion is still motion.

Very true. And zero point motion seems to not have any.

The thermal fluctuations are still random fluctuations.

True. A system with a unique ground state has no thermal fluctuations
when it is in the zero temperature zero entropy quantum ground state.

Suppose the particles are fermions. At high temperatures this
doesn't make any difference.

Now extrapolate to zero temperature. If you do it wrong, you get
zero pressure. If you do it the right way, you get nonzero pressure
at zero temperature. This has to do with the motion of the particles
at zero temperature. If there were no motion, there would be no
pressure. If there were no kinetic energy, there would be no
pressure.

Not necessarily. In general it has to do with just energy, not
specifically kinetic energy. But in the special case of a system which
is in a box with infinitely hard impenetrable walls rather than the
more general and realistic case having partially soft partially
penetrable walls, the energy does indeed happen to be entirely kinetic.
But the
(kinetic) energy and every other observable is time-*in*dependent when
the system is left undisturbed. The pressure is a rate of potential,
i.e. possible, virtual work that would be done on the system per unit
wall area if the location of the walls were to be virtually
infinitesimally displaced. For such a virtual displacement the
particles themselves in the quantum ground state remain in their
respective occupied single particle states adiabatically as the walls
are virtually displaced. The virtual displacement virtually changes
the energy spectrum of the single particle states in the system. The
total energy of the system is th e sum of
the changed energies of the occupied single particle states. As the
individual energies change the total sum value changes under the
virtual displacement. The rate this total changes per unit virtual wall
normal displacement distance, per unit area of the walls is the zero
point pressure. But nothing in the system is doing any actually moving
when the system is left undisturbed. And when it *is* disturbed by
having its walls displaced, the only thing about the system that is
changing is the individual particles just have their energies
adiabatically and trivially track, in real time, that imposed by the wall
displacement.

Indeed, if there were no kinetic
energy, there would be no energy at all, since the PE for this system
-- ideal gas in a box -- is zero under all conditions.

The zero mean PE is only *if* the walls are required to be perfectly
impenetrable rather than have any elastic give. Otherwise both kinetic
and potential energies contribute to relevant energy, which is just the
total energy of the states involved.

The pressure at zero temperature is sometimes called 'degeneracy'
pressure, but really it's not any different from any other kind of
pressure. It's just pressure. The quantum fluctuations are random
fluctuations, and they are not in any fundamental way different from
thermal fluctuations. Thermal and quantum are different asymptotes
on the *same* fundamental curve
http://www.av8n.com/physics/degeneracy.htm#fig-qho

The curve is a graph of total *average* equilibrium energy for a SHO
mode in a fixed temperature heat bath. It is not a graph of any
fluctuations per se, at all, let alone both thermal and quantum ones.
At the zero temperature limit of the graph there are neither thermal
nor quantum fluctuations in the energy. The entropy is zero there,
after all, and the ground state energy is unique with a well-defined
*non*-fluctuating nonzero value.

The zero-temperature pressure is well known in neutron stars, in
nuclei, and in metals. It is not a matter of opinion.

Of course. But that doesn't mean a system in its ground state has
anything about it that is actually moving in some way when the system
is left to itself. Pressure is simply not the same as motion.

Anybody who doesn't believe me is invited to do the calculation.
It's not a tricky calculation. Please let us know if your answer is
different from mine.
http://www.av8n.com/physics/degeneracy.htm

I'm sure you know how to properly calculate degeneracy pressure. But
no matter what its correct value happens to be that doesn't mean there
is any motion in the system possessing that degeneracy pressure.
Pressure is not the same as motion.

======================

Here's yet another line of reasoning that leads to the same answer.

Consider an atom. Better yet, consider a harmonic oscillator
consisting of an electron attached to a proton by a spring.
This is hard to build, but it's conceptually simple.

QM gives us a position operator for the electron. Also a velocity
operator. Also an acceleration operator. Also acceleration squared.
The expectation value of acceleration squared in the atom (aka
oscillator) is nonzero. Therefore in accordance with the Maxwell
equations, the thing will radiate. Even in the zero-temperature
ground state, the thing *must* radiate.

1) Note this is not an isolated system. It is a charged oscillator
interacting with the external Maxwell field.
2) The velocity operator for the non-interacting system is just the
momentum operator divided by the mass. And the acceleration operator
is just the position operator multiplied by the negative spring
constant over mass ratio. The squares of these operators are
proportional to the kinetic and potential energies respectively. These
squared operators have nonzero expectations because there are nonzero
probabilities for multiple different outcomes for a position
measurement and for a momentum measurement. I agree the
distributions
for potential position measurements and for potential momentum
measurements have nonzero variances.
3) That doesn't mean the system is radiating in its ground state,
Maxwell's equations notwithstanding.

Having uncertain values in a quantity is not the same as having
time-dependent values. For me (at least) motion requires having
time-dependent values, not just uncertain values.

Now things are getting interesting, because we know that on average
there must be no net radiation coming out of the ground state. Note
the contrast:
-- There must be radiation coming out.
-- There must be no average net radiation coming out.

This is not a contradiction. It is not a paradox. It is easy to
understand as soon as you realize that the atom is in equilibrium
with the EM field ... and the field has zero-point fluctuations of
its own. This works just like thermal equilibrium at any other
temperature: some thermal radiation goes out, and on average the
same amount of thermal radiation goes back in.

That's one way to look at it. One could also say there just isn't any
radiation going in or out when the system is in its ground state,
Maxwell's equations and quantum fluctuations in the value of quantum
variables notwithstanding. There is certainly no measureable radiation
going in or out.

This is super-easy to see if you analyze the system using something
resembling position operators and voltage operators.

In contrast, you will never see it using the photon-number basis,
because those are monochromatic. They have zero spread in frequency
and correspondingly infinite spread in time,

The oscillator system is also monochromatic, i.e. one frequency mode,
when it is considered by itself apart from the Maxwell field.

so they
average out all the fluctuations.

But measuring the radiation supposedly going in or out would require
measuring at least one photon's worth of radiation in transit. That is
not going to happen.

Insisting on analyzing the
zero-point motion in the photon-number basis would be like starving
to death at a banquet because they have run out of your favorite type
of pickle. I concede that one particular dish is empty, but
everywhere else you look there is plenty of food, plenty of evidence
for quantum fluctuations.

I don't think the other frequency dishes are relevant. If the
background EM field is a vacuum it has no photons at any and all
frequencies--including the frequency of the oscillator. If there *are*
occupied by photon modes at different frequencies the external field is
actually driving the oscillator off-resonance and the system is not
being left to itself in its ground state. I believe tickling the
system from the outside with different frequencies is cheating when it
comes to looking for motion in the system in its ground state.

In short, my point is a point about words, not physics. Physical
arguments won't affect that. My point is that zero point 'motion'
ought to be called that if something is actually moving, i.e. changing
in some way over time. It ought not be called that if nothing is
actually moving in any measureable way. I'm perfectly happy with zero
point energy and degeneracy pressure. But they don't demonstrate any
actual motion of anything.

BTW, JD, what did you think of my point about phase space areas
corresponding to pure quantum states?

David Bowman
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