NetNews Usenet Archive 1992 #27/NN_1992_27.iso/spool/bionet/infothe/987

Internet Message Format | 1992-11-17 | 3.4 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: 18 Nov 92 20:12:34 GMT
Sender: [e-mail]
Distribution: bionet
Lines: 103


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

> The other thing that proved to be kindling was my definition
> of temperature as average energy per state.  Here the disagreement
> is more interesting:
>
> >> 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.
>
> That's an interesting point.  Could you explain this a bit further?
> The correspondence isn't immediately obvious to me...

Okay... When you maximise information entropy with an energy constraint,

<E> = \sum_i p_i E_i

in addition to normalisation, you naturally obtain the Canonical
distribution,

p_i = 1/Z exp(\beta E_i),

where \beta is the Lagrange multiplier corresponding to the energy
constraint, and Z is the partition function. We can write the expected
energy as

<E> = -\frac{\partial}{\partial\beta} \log Z

The information entropy of this distribution is

S_I = \log Z(\beta) + \beta <E>

Now the E_i may depend on extrnally measurable parameters, such as volume,
magnetisation, stress, etc. So for each parameter q_r, we have a
generalised force,

F_r = <\frac{\partial E}{\partial q_r}

    = \sum_i p_i \frac{\partial E}{\partial q_r}

    = \frac{-1}{\beta}\frac{\partial}{\partial q_r} \log Z

(Almost there, guys) Now if we write the differential of the information
entropy,

dS_I = d\log Z(\beta,q_r) + \beta d<E> + <E> d\beta

     = \frac{\partial\log Z}{\partial\beta} d\beta
        +\sum_r \frac{\partial\log Z}{\partial q_r} dq_r
        +\beta d<E> + <E> d\beta

     = -\beta\sum_r F_r dq_r + \beta d<E>

and rearrange, we get:

d<E> = \frac{dS_I}{\beta} + \sum_r F_r dq_r

     = \frac{dS_I}{\beta} + dW

We are now in a position to identify this as the first law, and the first
term as heat,

dQ = TdS,

so \beta = 1/kT identifies with the thermodynamic temperature, and S=kS_I
identifies with the thermodynamic entropy.

>
> > 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)
>
> My definition was E per state, not per particle.

Point taken, but it still doesn't work, and the counterexample is still the
Fermi-Dirac case.
>
> In any case, the REAL test of which temperature definition is
> correct must be settled by checking which one gives the correct
> direction of heat flow when two systems of different temperature
> are "weakly coupled".  I would think that the average energy per
> state would equalize between the two---are you claiming that the
> exponential decay rates of the energy distribution equalize
> instead, in the case when these two definitions disagree?
>

Given the correspondence with the Thermodynamic temperature just
demonstrated, yes.

Kevin Moore,

[e-mail]


Typographcal errors notwithstanding, of course.