Re: macroscopic vs microscopic degrees of freedom

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

John Mallinckrodt asked how other phys-L-ers feel about the
debate between him and John D. I suspect that many are learning
from it; I certainly do. But I would prefer not to call the F*dx
the pseudowork. John M does not care about words, only about
the need to distinguish. Therefore he would probably agree to
preserve traditional terminology of introductory course and to
use the term pseudowork in more advanced courses. Rick's
concerns (see his last message) are important.

How would I answer the question John M asked again (see below)?
I would say that this question can not be answered unless the force
profile (the magnitude of F vs the distance x between the blocks) is
provided. Knowing the interactive profile one would calculate the
work done on each block as an integral of F*dx. What is wrong
with this?
Ludwik Kowalski

John M. wrote:

> 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