NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/bionet/infothe/977

Internet Message Format | 1992-11-15 | 4.7 KB

Path: sparky!uunet!biosci!buc.edu.au!kjm
From: [e-mail] (Kevin Moore)
Newsgroups: bionet.info-theory
Subject: Re:  A Mathematical Fireside Chat at 300K about 0K
Message-ID: <[e-mail]>
Date: 16 Nov 92 14:08:15 GMT
Sender: [e-mail]
Distribution: bionet
Lines: 100


In article <[e-mail]>
[e-mail] (Paul Burchard) writes:

> The energy of the system will be distributed among its states in some way.
> Let p(E) be the distribution (where E = energy per state).  The simplest
> measure that can be derived from this energy distribution is the
> *average* energy per state---this we call the "absolute temperature" T
> (working in suitable units).

WRONG! Consider the Fermi-Dirac distrubution to see the error with this.

> Now if all we know about a system is its temperature T, what can we
> say about p(E)?  Well, not forgetting our other assumption, that E>0,
> we can use information theory to make the *least presumptive* guess
> for p(E).  We do this by maximizing the information-theoretic *entropy*
> of the probability distribution, constrained by our two assumptions.
> In this way we find that our best guess is the exponential distribution
> p(E) = (1/T) e^{-E/T}.

That's p(E) = (1/Z) e^{-E/kT} where Z is a normalisation constant,
otherwise known as the partition function (or functional, in the
non-equilibrium case).

             An ideal system that actually conforms to this
> generic energy distribution we will call a "heat bath".
>
> Note that T also shows up as the exponential decay rate of p(E).  This
> suggests an alternative definition of temperature, which allows the
> possibility of "negative absolute temperature".  We ignore this definition.

Why? That T corresponds to the integrating factor in dQ=dS/T. It therefore
corresponds to the thermodynamic definition.

    [deletia]

> Our conclusion?  Temperature implies an upper bound on (physical)
> entropy, and under physically correct assumptions, this upper bound
> goes to zero as the temperature goes to zero.  Therefore,
> zero temperature implies zero entropy.

I'll go along with this.

> But...you may be troubled by the following elements of the argument:
>     1. Saying that E is always positive is tantamount to making
>         E an absolute, rather than relative, quantity.
>         This flies in the face of classical mechanics.

E does not have to be positive for statistical mechanics to work. You
simply get different values for Z with different zeroes of potential
energy.

>     2. In order to make sense of the limiting maximum entropy
>         as T goes to zero, we had to assume discreteness
>         of energy levels.  Classically, it's trouble again.
> In fact, I'm skeptical that we can make classical sense of entropy at
> 0K.  The entropy bound's divergence to minus infinity reflects the
> infinite information difference between a continuous and a discrete
> probability distribution.  It might still be possible to renormalize
> these calculations somehow, but that's a slippery business in itself.

Yep. You use $\int \rho\log\frac{\rho}{m} d\tau$, where $m$ is the Lebesgue
measure on phase space $\tau$. The $m$ is important, because it makes the
information entropy invariant under choice of parametrisation of phase
space. Klir, the fuzzy set person, (unforgivably) sets up a straw man by
leaving out the measure when criticising the maximum entropy procedure.


> Quantum physics addresses both problems.  First of all, it gives us
> discrete energy levels, at least for finite systems.  Actually, all
> we really need is that there be a finite energy gap at E=0.  This is
> a very mild assumption quantum mechanically.

> More substantially, QM typically gives us an absolute zero energy level.
> This zero level *need not* be the ground state of the system; it is
> sensible (as sensible as QFT gets :-) to ask if the ground state energy
> vanishes.  The most familiar example of this is the quantum harmonic
> oscillator, where the ground state energy is  hw/2  (h = Plank's h-bar;
> w = angular frequency).  Notice that if the ground state energy does not
> vanish, then the system cannot get to 0K, purely by definition.

As I said earlier, the average energy definition is a BAD one. In fact it
only works for classical particles, with no exchange interaction. All the
particles in the ground state is 0K. ($\beta=1/kT$ is infinite) The quantum
mechanical reason for the third law is the energy-time uncertainty
relation. In order to establish 0K, you have to constrain the energy state
for an infinite time.

> So now I'll go warm my hands by the resulting flamefest!!!
> I vote for 0 entropy at 0K!!!

> P.S.  Please criticize my *definitions* first, before flaming
> my conclusions!

I did, and I hope you're satisfied. :-)

Cheers,
Kevin.

[e-mail]