Resolution to Gibbs' entropy paradox?

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SUMMARY

The discussion centers on resolving Gibbs' entropy paradox by proposing that particles can be distinguishable, contrary to traditional assumptions. This approach suggests that the total number of microstates should account for distinguishable particles, leading to the conclusion that the entropy of a classical ideal gas is extensive. The solution involves multiplying the phase-space volume by the combinatorial factor "A choose N," resulting in the standard entropy expression plus an additional term that has no observable effects. This perspective aligns with Jaynes' interpretation of entropy as a function of probability distributions.

PREREQUISITES
  • Understanding of Gibbs' entropy paradox
  • Familiarity with phase-space volume concepts
  • Knowledge of combinatorial mathematics, specifically "A choose N"
  • Basic principles of statistical mechanics and thermodynamics
NEXT STEPS
  • Research Jaynes' approach to entropy and its implications on information theory
  • Explore the concept of distinguishable particles in statistical mechanics
  • Study the mathematical derivation of phase-space volume and its applications
  • Investigate the relationship between entropy and information content in physical systems
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Physicists, thermodynamicists, and researchers in statistical mechanics seeking to deepen their understanding of entropy and its foundational concepts.

maline
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It seems to me that Gibbs' Paradox (that the entropy of a classical ideal gas, calculated by phase-space volume, is not extensive) can be resolved without assuming that particles are indistinguishable.
Suppose instead the opposite: that particles are distinguishable, meaning that each one can in principle be identified- imagine a minuscule serial number stamped on every molecule. Now this should apply not only to the system under consideration, but to the universe as a whole- the serial numbers run from one to A, where A is the total number of molecules (of a particular type) in the universe.
This immediately implies that specifying the position and momentum of each of the N particles in our box does not fully determine the microstate of the system! We must also specify which, out of the A molecules in existence, are in fact the N ones in the box. The total number of microstates should include all such possibilities.
This means multiplying the phase-space volume by "A choose N", that is, A!/(N!(A-N)!). Since A>>N, the factor A!/(A-N)! tends to AN. Thus we are left with the desired factor of 1/N!, giving the standard (extensive) entropy, plus a constant contribution N log(A) that is also extensive and (I think) has no observable effects.
 
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maline said:
It seems to me that Gibbs' Paradox (that the entropy of a classical ideal gas, calculated by phase-space volume, is not extensive) can be resolved without assuming that particles are indistinguishable.

You seem to have stumbled upon Jaynes' solution, which results in a connection between information and entropy:

http://www.santafe.edu/media/workingpapers/07-08-029.pdf
 
Where in the book is this?
 
maline said:
Where in the book is this?

Chapter 5.
 
I don't see my argument there. All I see is the assertion by Gibbs that identical particles must be treated as indistinguishable. There is a completely different conception of entropy there, due to Jaynes, in which the entropy is a function of a probability distribution that we assign, rather than of the physical system per se. I am working with the original phase-space volume concept, and simply pointing out that if particles were distinguishable, then that gives us more possibilities for "different" microstates.
 
maline said:
I don't see my argument there.

Then keep reading, that includes searching the literature- and there's a lot of it. There's a clear connection with information content and distinguishable particles.

Edit- how about this: a thought experiment. Consider a box partly full of spheres; when you look inside you see a partition separates one half, containing green spheres, from the other half which has red spheres. However, when I look in the box, because I have special glasses that only transmit luminance values, I see a partition separating equal numbers of grey spheres.

Now we remove the partition and shake up the box. When we look inside and compare the final state to the initial state, do we initially report the same or different increase of entropy? (and explain your answer)
 
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Andy Resnick said:
how about this: a thought experiment. Consider a box partly full of spheres; when you look inside you see a partition separates one half, containing green spheres, from the other half which has red spheres. However, when I look in the box, because I have special glasses that only transmit luminance values, I see a partition separating equal numbers of grey spheres.

Now we remove the partition and shake up the box. When we look inside and compare the final state to the initial state, do we initially report the same or different increase of entropy? (and explain your answer)
Yes, this is a standard statement of the paradox. The answer is well- known: there is an increase in entropy, but the guy with the glasses will not know this. But what exactly is his error? According to the standard account, the problem is that he wrongly thinks the balls are indistinguishable, and therefore undercounts the microstates in the final situation. I am suggesting the opposite: if the balls are in principle distingushable- and macroscopic balls certainly are- then his glasses cause him to overcount the microstates initially, by neglecting the color constraint that limits which balls can be on each side.
 
HPt said:
I published this solution 2010 in Journal of Statistical Physics http://www.springerlink.com/openurl.asp?genre=article&id=doi:10.1007/s10955-010-0077-7 and, more concisely, 2014 in European Journal of Physics http://dx.doi.org/10.1088/0143-0807/35/1/015023
Unfortunately, the solution is still widely unknown.
This is great! It restores my faith in the existence of common sense in physics. I especially love this quote:

"Secondly, the resolution above implies that the mere concept of distinguishable identical particles is at odds with thermodynamics [12]. In the opinion of the present author, such a connection between these otherwise unrelated subjects would be rather surprising, to say the least."
 
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  • #10
maline said:
Yes, this is a standard statement of the paradox. The answer is well- known: there is an increase in entropy, but the guy with the glasses will not know this. But what exactly is his error? According to the standard account, the problem is that he wrongly thinks the balls are indistinguishable, and therefore undercounts the microstates in the final situation. I am suggesting the opposite: if the balls are in principle distingushable- and macroscopic balls certainly are- then his glasses cause him to overcount the microstates initially, by neglecting the color constraint that limits which balls can be on each side.

I would caution you against using 'error', 'wrongly', etc, because you are making some assumptions. The basic point, the essential starting point, is only that we initially report different values. Our values become equal once you provide me with additional *information*.

There's a deep principle here, because it fundamentally goes to the idea of 'hidden variables' and how we can decide if we have complete information about a system. Because surely you would allow that perhaps you did not notice that there were actually *4* colors instead of 2, so your calculation is 'wrong' as well. This has led some to wonder if entropy is not an objective, external, feature of the universe but instead an anthropomorphic, subjective, one.

Since there are no hidden variables, we instead postulate that the difference in reported values is quantified by information: 1 bit of information has entropy k ln(2).

http://www3.imperial.ac.uk/pls/portallive/docs/1/55905.PDF
 

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