Re: Correct Answer

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Brian Whatcott asked:
> Actually, because any other position does not minimize potential energy,
> to me this sounds like a very counter-intuitive example of
> 'Least Action'.
>
> Is it?

It is not so much an example of least action as it is an example of maximum
entropy.  The reason that macroscopic systems come to equilibrium by
dissipating the macroscopic energy in their macroscopic degrees of freedom is
that the second law requires that as time goes on the entropy associated with
the system *and its surroundings* will continue to increase until it can no 
longer do so.  If the energy in the macroscopic modes is spread out and
shared (mostly) equitably among the myriads of microscopic degrees of freedom
then the overall entropy is increased.  (It takes more information to specify
how the energy is shared when it is widely distributed than if all the energy
is concentrated in a few macroscopic degrees of freedom).  The result of this
sharing is that the remaining share of energy that the macroscopic modes get
to keep is approximately the same as each of the other degrees of freedom
get, i.e. about kT worth above the minimal energy configuration (recall the
Equipartition of Energy theorem).  At room temperature this means that in
equilibrium the potential energy of the oscillation modes averages at about
(1/2)*k*T = 2.06 x 10^(-21) J above the minimal potential energy value, and
the average kinetic energy in this mode is also about 2.06 x 10^(-21) J above
the state of rest.  Thus, for all practical purposes the system continues to
loose energy from its macroscopic modes until they are in a state of rest at
the minimal potential energy configuration.

David Bowman
dbowman@gtc.georgetown.ky.us