Unifications of entropies.
It is an offensive manner in these groups to humiliate
famous scientists showing their errors, actual or
supposed. I will continue in this habit. Even
antieinsteinists will be satisfied.
A Nobelist boasted before his peers that he was the
first, who used partitions
to solve physical problem. Nobody corrected him. And not
from politeness. It seems
that one important priority is completely forgotten,
despite that the treatise, where partitions appeared is
cited also on internet, and the problem is (or when I
studied, was) a topics of lectures for beginners.
Boltzmann, trying to explain his formula for
thermodynamical entropy, published (1877) the following table:
(7,0,0,0,0,0,0)
(6,1,0,0,0,0,0)
(5,2,0,0,0,0,0)
(4,3,0,0,0,0,0)
(5,1,1,0,0,0,0)
(4,2,1,0,0,0,0)
(3,3,1,0,0,0,0)
(3,2,2,0,0,0,0)
(4,1,1,1,0,0,0)
(3,2,1,1,0,0,0)
(2,2,2,1,0,0,0)
(3,1,1,1,1,0,0)
(2,2,1,1,1,0,0)
(2,1,1,1,1,1,0)
(1,1,1,1,1,1,1)
These are all partitions of 7 into 7 parts.
p(7) represent, according to Boltzmann, all possible
distributions of 7 quanta (of energy, sic, long before
Planck) between 7 molecules. Boltzmann called partitions
complexions. He conjectured that molecules, changing
their energies by collision, at last reach a partition
giving them largest freedom.
I have written the rows of the table as row vectors, with
brackets and commas.
Instead complexions we will use the term "orbit". The
first partition in the table gives 7 distinguishable
possibilities how to place 7, the last one is unique.
It should be an elementary problem to find solution.
Apply S(n) group (n = 7), eliminate undistinguishable
permutations, which permute the same values of energy.
The volume of orbits is determined by a polynomial coefficient
n!/(Pi)nk!
k = 0,1,2...infinity.
Boltzmann conjectured, that logarithm of volume of
individual orbits is proportional to thermodynamical entropy.
Applying the Planck constant for quanta of energy k, and
Avogadro number for number of molecules, the exact
factorials were replaced, using Sterling formula as
ln n! = nln(n),
and the famous formula obtained
H = -k(Sum)n(k)/nln(n(k)/n).
It was Boltzmann's tragedy, that his approach was not
understood and that it was completely forgotten.
The comedy continued (nobody knows, what entropy is...).
The theory of communication was promoted as the theory of
information and its axiom replaced dubious Boltzmann definition.
Shannon used entropy as the measure of uncertainty,
actually as an evaluation, how many messages can be
formed from the given set of symbols. He used an example
with 8 symbols, to make calculations easy. I hope that
you master even 7 symbols:
(a,a,a,a,a,a,a)
(a,a,a,a,a,a,b)
(a,a,a,a,a,b,b)
(a,a,a,a,b,b,b)
(a,a,a,a,a,b,c)
(a,a,a,a,b,b,c)
(a,a,a,b,b,c,c)
(a,a,a,b,b,b,c)
(a,a,a,b,b,c,c)
(a,a,a,a,b,c,d)
(a,a,a,b,b,c,d)
(a,a,b,b,c,c,d)
(a,a,a,b,c,d,e)
(a,a,b,c,d,e,f)
(a,b,c,d,e,f,g)
It is not customary to divide consecutive symbols in
texts (messages, use any term you like, I employed
strings). Texts carry information, they are information vectors.
Moreover, it is practical, when we compare both tables.
There exists one to one correspondence between them, when
we use alphabetical index. But when we make symmetry
operations with the symmetry group S(m), m = 7, we get
another result than before. The first row gives only the
identity, the last one all 7! permutations.
It is possible to let act both groups simultaneously.
Thus both operations are multiplicative.
I would be underestimating the intelligence of the
readers of these groups, if I continued with explanations.
Excellently, even Newton can be blamed, in his formula
only one coefficient appears.
Unfortunately, at least for me, the product of two
polynomial coefficients can be found in a handbook of
statistics. Even I am not the first one in interpretation
of entropy as a measure of symmetry.
A hint for antieinsteinists: Differentiate the result of
the generating function according to values of one vector.
Literature:
Boltzmann, L. Wien. Ber., 76, 373 (1877).
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