Suppose we have a fair lottery drawing machine where you have a container of numbered balls that is rotated many times such that interaction of the balls with themselves and the container produces a random distribution of the balls. We can say that this distribution of balls has a certain entropy, call it $E$. Now we have some kind (the details are unimportant for the question) of extracting mechanism that extracts balls from the container, there is an entropy associated with this extractor , call it $E-prime$. Would it be correct to say that the resulting lottery picks are the result of the superposition of entropies $E$ and $E-prime$ or the product of both entropies?

  • $\begingroup$ Why does the extractor device have an entropy? $\endgroup$ – boyfarrell Oct 30 '18 at 22:07
  • $\begingroup$ @boyfarrell: Because the extractor picks the balls from any point inside the container, a probability space if you like, not just selecting a ball from a single point as the balls go passing by. Should I edit my question to add that feature? $\endgroup$ – William Hird Oct 30 '18 at 22:13
  • $\begingroup$ OK. I normally think of associating an entropy to a distribution. I think that the extraction device is just a channel. You could think about this equation as an analogue, en.m.wikipedia.org/wiki/Landauer_formula which is a channel connecting two distributions. But really here you only have one distribution. $\endgroup$ – boyfarrell Oct 30 '18 at 22:19
  • $\begingroup$ @boyfarrell: I think there are two distributions, the distribution of the balls at the moment a ball is selected and the distribution of the extracting locations as plotted against time. $\endgroup$ – William Hird Oct 30 '18 at 22:43
  • $\begingroup$ I think there is one distribution of particles defined by phase space of position and momentum values (x, y, z, px, py, pz). The extractor samples a single phase space point because it is a physical mechanism, say at (xe, ye, ze) and extracts balls with any momentum. If the balls are ergodically distributed, then it does not matter the location of the extractor or even if it is at a constant location with time. $\endgroup$ – boyfarrell Oct 30 '18 at 23:35

Would it be correct to say that the resulting lottery picks are the result of the superposition of entropies E and E-prime or the product of both entropies?

TLDR: the entropy of the total system is $E+E’$

Full answer: the entropy of a system is related to the number of micro states that a system could be in for a given macro state. Now, it is a little odd to discuss macroscopic machines in terms of macrostates as though we cannot observe the relevant details of the machine. But nevertheless we can discuss macro and micro states in general.

So for any system the entropy is related to the number of microstates by $E=k \ln(\Omega)$. Assuming that the two systems are independent and that each microstate is equally likely then the number of possible microstates in the combined system is the product of the number of microstates in the two sub systems. I.e. for every microstate of one subsystem the other could be in any of its possible microstates.

Then, the entropy of the combined system can thus be directly calculated as $k \ln(\Omega \; \Omega’) = k \ln(\Omega) + k \ln(\Omega’) = E + E’$

  • $\begingroup$ Nice answer. So how would this relate to showing that such a system has maximum entropy ? $\endgroup$ – William Hird Nov 3 '18 at 4:11
  • $\begingroup$ If each sub system had maximum entropy then the entropy of the combined system would be maximum and equal to the sum of the two maximum entropies. If one system did not have maximum entropy then the tendency would be to exchange entropy so as to maximize the total entropy. $\endgroup$ – Dale Nov 3 '18 at 12:00

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