Re: ENERGY WITH Q

["RAUBER, JOEL" <JOEL_RAUBER@SDSTATE.EDU>]

> > I do like the spin glass example because, as you say, it is easy
> > to calculate the absolute entropy.  But I don't yet see why that
> > absolute value is *important*, that is, why one could not add an
> > arbitrary constant to it without changing any measurable
> > thermodynamic results.
>
> Well, let's work it out.
>   1) Suppose that in addition to the R log 2 of molar entropy
> that we know
> about (one bit per nucleus), suppose we had an additional 17 bits per
> nucleus of "secret entropy".
>   2) Assume antiprotons have just as much "secret entropy" as regular
> protons.  (It wouldn't make sense for them to have negative entropy.)
>   3a) The foregoing suppositions are inconsistent with the observed
> thermodynamics of proton-antiproton annihilation. QED.

Could you elaborate on 3a)?

The way I see it (perhaps incorrectly),

We have a system consisting of a proton and an anti-proton.

Following John M.'s suggestion and your example we have

initial Entropy = S_i = 34 + (the "correct" absolute value for the proton
anti-proton pair) bits

The process of the proton and anti-proton annihilation occurs.  The 2nd Law
says that

Delta S >=0

so I conclude

Final entropy = S_f = S_i + Delta S = 34 + (entropy of the two resulting
photons, based on calculating "correct" absolute values).

Replace 34 with any fixed number of your choosing and we still have the same
value for Delta S.

So how was it inconsistent with the thermodynamics of the annihilation
process?

Note:  I'm not adding 17 units per nucleus; I'm adding 34 to the system
under consideration!  I don't think I'm required to think of it as 34 per
nucleus?

I agree with you and John that in these simple examples it is rather
pointless to add such a constant, but . . .


Joel R.