Re: How to calculate entropy of particles?

On Feb 15, 10:12 am, Rick Giuly <rgiuly.gr...@xxxxxxxxx> wrote:
On Feb 14, 11:40 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:





On Feb 14, 9:22 pm, Rick Giuly <rgiuly.gr...@xxxxxxxxx> wrote:

Hello all,

I'm working on a molecular dynamics simulation based on Lennard-Jones
potential.  Each particle has a position, a mass, and a velocity.

I know that temperature is the average kinetic energy of all the
particles, so temperature is no problem to calculate.

But, how would you calculate the entropy of the particles?

S = k ln(number of states)
dS = dQ / T

Any help is appreciated.

-Rick Giuly
If replying by email please use: rgiuly at ucsd dot edu

reference:http://en.wikipedia.org/wiki/Lennard-Jones_potential

The problem I have with S = k ln(number of states) is that it seems
like the number of states is nearly infinite, since there are multiple
particles and each can be at any location in space.

 dS = dQ / T
This formula tells me something about how the entropy would change
when heat is added but I want to calculate the entropy as a function
of the position and velocity of all the particles at an instant in
time.

So I'm still not sure how to calculate the entropy (as a function of
the position and velocity of all the particles at an instant in time).

Annoying people like to say that entropy is a measure of our
ignorance. Specifically, it's a measure of how many microstates are
compatible with a given macrostate: the "macrostate" is a description
of our knowledge about the system (temperature, pressure, volume...
etc.).

Given a particular microstate, I don't see any necessary meaning to
its "entropy", although for extreme fluctuations, we might mean "the
entropy of the resulting restricted set of microstates if we
partitioned the system at that instant".

Example: by random fluctuation, more molecules will in general be in
one half of a box than the other half. If we lowered a shutter at
that instant, this fluctuation would be preserved in macrostate as a
density and pressure differential between the sides. If we calculated
the entropy of this resulting pair of systems, it would be lower than
the original unidivided system: i.e., the measure of the volume of
microstates it could occupy would be smaller. In this sense we might
say "the microstate showed an entropy fluctuation". However, the
value of this fluctuation would depend on how we partitioned the
system when we took the snapshot, lest you think the idea is
unambiguous.

Another problem seems to be the application of S = k ln(omega) to
cases where the possible states are "nearly infinite" -- i.e.,
continuous. Presumably omega is replaced by a volume in phase space.
The first problem is more fundamental.

Entropy is not like "mass" or "energy", which have a well defined
value for a microstate: it's a property of macrostates.
.

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