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Re: first law of thermo



"Carl E. Mungan" wrote:

F = m dv/dt => F dot dr = m v dot dv

Integrate both sides to get desired result W = delta(K). Add
subscript "com" to W and subscript "TR" to K if you prefer (but you
said this wasn't an argument over nomenclature).

As for terminology, it would be helpful but not
necessary to employ subscripts or some such, for
clarity.

But with or without any particular terminology, the
usefulness of the foregoing calculation is very limited.
-- this W does not represent change in the total kinetic energy
-- this W does not represent change in the thermal energy
-- this W does not represent change in the nonthermal energy.

More generally, I have no idea what this W is good for.
As far as I can tell, it is neither necessary nor sufficient
for deriving and/or explaining thermodynamics. As far as
I can tell, it is a mathematical curiosity which is at best
little more than a waste of time, and more commonly is a
distraction, deflecting attention away from the real,
useful thermo ideas.

For more discussion of this point, see

http://www.monmouth.com/~jsd/physics/momentum-squared.htm#eq-mom-squared


Hint: I've got a pocketful of counterexamples, so
any such derivation is going to be quite a rare bird.

Fire away.

The original rotational version of Rumford's experiment
imparts no center-of-mass KE to anything. Yet there
is work (according to the commonsense definition of the
word, which agrees with the technical definition used
by every research physicist I've discussed it with) and
there is production of heat (according to same sort of
definition).


=====================


VIEWPOINT 1: VIEW THE ENTIRE SYSTEM (two blocks, spring, and table)
AS ONE BULK WHOLE.
What I'd say is that the interaction with the table has converted
*one form* of internal energy (namely macroscopic rotational and
vibrational) to *another form* of internal energy (namely microscopic
kinetic and potential).

OK.

Bulk energy refers purely to translational KE
of the system's com BECAUSE OF OUR VIEWPOINT. This is zero both
before and after the interaction, so W-K correctly predicts 0=0.
Also, there are no external forces on the system, so FLT correctly
predicts 0=0. This view is thus not very helpful. Let's go on.

The "bulk" property of energy is a red herring.
It has no significance to thermodynamics.
It is a waste of class-time to introduce the concept.
What matters is entropy or the lack thereof.

VIEWPOINT 2: BREAK THE SYSTEM INTO MACROSCOPIC PARTS.
Suppose we're not happy with calling the bulk rotational and
vibrational "internal energy". Fine, I can break the system into
three parts: the two masses and the table. If you will permit me, I
will take the spring to be massless (ideal) so it doesn't count as a
part.

Massless is what I intended, so no problem there.

In that case, we have mechanical energy = sum of K and U for
the parts = mv^2 + kx^2/2 where I assumed for simplicity that each
block had the same mass m and speed v and the "centrifugal" stretch
of the spring is x.

Fine. Clear.

And we have initial internal energy = E1 (which
reflects the temperatures and phases of the blocks and table). W_ext
= delta (E_mech) + delta (E_int) now predicts 0 = -mv^2 - kx^2/2 + E2
- E1 where E2 is the final internal energy and where I assumed spring
relaxes fully.

Ah, but what about viewpoint 3: Suppose I break the system
into two parts, not four:
-- table, and
-- blocks+spring

Then the Work/KE theorem predicts _zero_ which is not
the right answer.

I don't trust a formalism which produces the right answer
if you know the right answer and produces lots of wrong
answers just as easily.