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Re: Entropy -- sorted -- side bar

John --

It wasn't a typo. A standard as-purchased deck as two jokers, which is why
I wrote 54 factorial.

OK, but for those of us who are familiar with casino decks, there are no
jokers or advertising cards


That's right. Or, if you want to measure entropy in units of bits, use
base-2 logarithms and lose the factor of k:
S = log_2(# of microstates)

And in particular 54! = 2.30844E+71
and log_2(54!) = 237.0638
unless my spreadsheet is lying to me.

Why do you drop the k???

Now as a side bar to this thread perhaps someone can calculate the
following -- another counting problem:

Consider a box with volume V which contains N identical ideal particles. If
it is assumed that each particle can occupy a single 6-d Heisenberg space
of DxDyDzDpxDpyDpz =~h^3 (or what ever you like) and that the total energy
of the system is E (Assume all other good things)

What is the entropy of this system?

One may peek at my First Law page to see how I have done this -- but I
worry that that derivation is incorrect. The worry is based in the state
counting.


Jim Green
mailto:JMGreen@sisna.com
http://users.sisna.com/jmgreen