Re: macroscopic vs microscopic degrees of freedom

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Sun, 31 Oct 1999, John Denker wrote:

> My notion of "conventional" work excludes microscopic random energy
> transfers.  These are just as "mechanical" as nonrandom transfers *if* you
> look super-closely, but still I claim they should be called heat whereas in
> a previous posting you seem to say should be classified as work not heat.

Yes, and I still prefer to look at it that way, but I understand that
others may view it differently.  There are no mistakes here, only
differences of opinion about terminology.

I am curious (seriously) about how you would classify the energy transfer
that takes place in the collision problem I posed the other day.  To
remind you, bodies of mass 2 kg and 1 kg approach each other both moving
at 1 m/s.  After the collision the 2 kg body is at rest (and, therefore,
the 1 kg body moves off at 1 m/s in a direction opposite its initial
velocity.) I later added the information that the internal energy of the 2
kg block rises by 1 J during the collision.

I *think* you would say that the 2 kg block has had -1 J of work done on
it (because its kinetic energy has been reduced) and that it has been
heated by +1 J.  I also think you would say that the 1 kg block has had no
work done on it and no heat added to it.  In any event, I'm quite
confident that you would be able to calculate other things like delta T's
and delta S's correctly given the necessary additional information.

In contrast, I would say that no heat is involved in this process for
either block. The works done would depend on the definition used for work.
I can give you consistent and useful formalisms in which the works
(calculated in the "lab frame") are as follows:

Formalism          Work on 2 kg block        Work on 1 kg block
   #1                     -1 J                       0 J
   #2                      0 J                       0 J
   #3                     +1 J                       0 J

Furthermore, I assure you that I would be able to calculate other things
like delta T's and delta S's correctly with the necessary additional
information.

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm