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Shannon Entropy and Origins of the Universe

  1. Jan 20, 2014 #1
    Hello Community,

    I have a question that I'm struggling to get clarification on and I would greatly appreciate your thoughts.

    Big bang theories describe an extremely low thermodynamic entropy (S) state of origin (very ordered).
    Question: Is the big bang considered to be a high or low shannon entropy (H -information) state?

    Here's why I question:

    S and H are related: S = the amount of H needed to define the detailed microscopic state of the system, given its macroscopic description.
    Thus: Gain in entropy always means loss of information.

    2nd Law of Thermodynamics: S always tends to increase for closed systems.
    Therefore: H is generally decreasing.

    This suggests a very high Shannon entropy state at the big bang.

    However Shannon entropy is very closely connected to the number of degrees of freedom, etc.
    i.e Wasn't there a lot less degrees of freedom around the time of the big bang? Also, wasn't the universe a lot smaller and given the max information density, didn't it have a much smaller max H?

    This suggests a very low Shannon entropy state at the big bang.

    I'm confused, can't find anything on this anywhere and would love some input.

    Blessings to you all :)
     
    Last edited: Jan 20, 2014
  2. jcsd
  3. Jan 21, 2014 #2
    Anyone out there?
     
  4. Jan 21, 2014 #3

    Chronos

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    That is the kind of question that has led researchers to conclude we need a different playbook to fathom the origin of the universe. One that we have been fiddling with for most of the past century is called quantum gravity. For a deeper discussion of Shannon entropy, this may be helpful; http://micro.stanford.edu/~caiwei/me334/Chap7_Entropy_v04.pdf, and, http://arxiv.org/abs/0712.0029 Entropy Growth in the Early Universe. There is some dissent on whether information is truly lost due to entropy as illustrated by the Hawking and Susskind debates over information loss in black holes.
     
  5. Jan 26, 2014 #4

    JesseM

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    Where have you seen H defined this way in a physical system? Shannon entropy deals with situations where you are communicating outcomes drawn from some probability distribution. If outcomes have different probabilities, H is a -1 times a weighted average of the logarithm of the probability for each outcome, weighted by the probability of that outcome--in an optimal coding scheme where more probable outcomes are assigned shorter bitstrings, this would be the same as the average length of the bitstring you get if someone repeatedly draws outcomes from this probability distribution, and on each trial they send a string of bits that tells you the outcome on that trial. If all outcomes are equally probable, the Shannon entropy reduces to just the logarithm of the number of possible outcomes.

    The formulas for physical entropy look identical, except for being multiplied by plus-or-minus Boltzmann's constant k. In a situation where different microstates have different probabilities (my recollection is that this is usually a scenario where the system is connected to some external reservoir that it can trade energy/volume/particles with, and that the system may be in different possible macrostates with different probabilities), the Gibbs entropy is -k times a weighted average of the logarithm of the probability for each microstate, weighted by the probability of that microstate. And the Boltzmann entropy for a given macrostate is k times the logarithm of the number of microstates associated with that macrostate.

    To think about the relation between Shannon entropy and physical entropy, I think it's easier to consider Boltzmann entropy. Suppose someone is measuring the exact microstate of a system whose macrostate you both already know. Here all microstates associated with that macrostate are equally probable, so the Shannon entropy is just the logarithm of the number of possible microstates they might find. We could imagine a large set of trials where they repeatedly measure the microstate of systems prepared in that same known macrostate, and communicate the result to you each time--the lower the entropy of the macrostate, the shorter the bitstring they would need to communicate the microstate to you on each trial (for example, in an extremely low-entropy macrostate where there are only 2 possible microstates, a 1-bit message would be sufficient to tell you the microstate each time...but with 4 possible microstates they'd need 2 bits, with 8 possible microstates they'd need 3 bits, etc.) So it seems that the Shannon entropy of this sort of "outcome"--a measurement of a microstate given a known macrostate--is just the physical Boltzmann entropy of the macrostate with Boltzmann's constant taken out, meaning that a gain in physical entropy is a gain in Shannon entropy, not a loss as you suggested.

    I guess an alternate situation would be if someone was randomly drawing microstates from the system's entire phase space of possible microstates, and then sending you a message only about the macrostate--in this case, since high-entropy macrostates would be far more probable, in an optimal coding scheme they would involve shorter bitstrings, whereas low-entropy macrostates would require longer bitstrings. But Shannon entropy is supposed to involve a weighted average of the length of all possible bitstrings you might get in an ensemble of trials, not the length of any specific bitstring, so this still doesn't quite seem to make sense of the claim that higher entropy states involve smaller Shannon entropy. Perhaps what you mean is that low-entropy macrostates have higher self-information--the length of the bitstring required to communicate them in an optimal coding scheme (with Shannon entropy being the expectation value for the self-information as shown here)--assuming that you are communicating messages about the macrostate only, drawing the system's state randomly from the set of all possible microstates.
     
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