1. Probability Grouping and Shannon's Entropy
- Author
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Ruggeri, Francesco R.
- Subjects
pressure ,Shannon's entropy ,state probability - Abstract
Shannon’s entropy is given by - Sum over i P(i) ln(P(i)) where P(i) is the probability for the state i. The question is: What is the state i? In particular, different choices of the states in question lead to different entropies. Traditionally, the states associated with a gas with no degeneracy in energy levels are those for which energy E is distributed among N particles, with each distribution carrying the same weight. The number of permutations of a specific { n(ei) } set is: N!/ Product over i n(ei) where Sum over i n(ei) = N and Sum over i ei n(ei) = E (or Eaverage for a sharp distribution) may map into different physical configurations with different pressures. For example, an arrangement of three e1’s followed by 4 e2’s may not be the same physical situation as 1 e1, 2 e2s, 2 e1s 2 e2s. In other words, there may be a pressure related reason for choosing states. If each distribution carries the same weight, then ln(1/#arrangements) = ln(Probability of the average scenario of N particles). For N large, n(ei) = Np(ei) where p(ei) is single particle probability for independent particles. Then the average probability corresponds to: Product over i (p(ei)) [to the power of Np(ei) ]. Ln(average probability) = Sum over i p(ei) ln(p(ei)) + ln(N). -Sum over i p(ei) ln(p(ei) is then the Shannon’s entropy of the gas for no degeneracy in energy levels. The degeneracy in energy levels also seems to assume an average n(ei) for each ei as we discuss in this note (following (1)) i.e. considering boson and fermion cases where again the idea of pressure related states appears. In (2) it seems that a different grouping of probabilities is made to define physical states. (2) suggests that single particle situations which have the same number of n(e1), n(e2), n(e3).. etc i.e. specific sets { n(ei) } such that Sum n(ei) = N and Sum over i ei n(ei) = E(average) should be grouped together i.e. have their probabilities summed. Thus all permutations of the same set of n(ei)’s are grouped into one probability and this is used to compute Shannon’s entropy i.e. Probability = N!/ Product n(ei)! Product p(ei) [to the power n(ei)]. No degeneracy in energy levels is considered and ultimately Maxwell-Boltzmann distributions are used for single particles. This then leads to a different entropy than using the p(ei)’s corresponding to the average picture (i.e. average of all the equally weight distributions of E over N). Arguments are made that this entropy allows for -ln(N!) to disappear as T->0 as all particles drop to the ground state. For S=- ln(N!/ Product over i n(ei)!) , T→0 also means that the MB distribution no longer holds and that n(ground state)=N and so S also tends to 0.
- Published
- 2023
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