What is Liouville's theorem all about - Thermodynamics

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SUMMARY

Liouville's theorem states that the volume of microstates in phase space remains constant under Hamiltonian flow. In the context of an ideal gas undergoing adiabatic free expansion from volume V to 2V, the confusion arises because the number of microstates appears to increase despite the theorem's assertion. The key insight is that Liouville's theorem applies only to reversible processes; thus, during irreversible processes like adiabatic expansion, the phase space volume is not conserved, leading to an increase in entropy and a greater number of macrostates.

PREREQUISITES
  • Understanding of Liouville's theorem in classical mechanics
  • Familiarity with thermodynamic concepts such as adiabatic processes
  • Knowledge of phase space and microstates in statistical mechanics
  • Basic principles of Hamiltonian dynamics
NEXT STEPS
  • Study the implications of Liouville's theorem in reversible versus irreversible processes
  • Explore the relationship between entropy and phase space volume in thermodynamics
  • Learn about Hamiltonian mechanics and its applications in statistical physics
  • Investigate the concept of microstates and macrostates in greater detail
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Students and professionals in physics, particularly those studying thermodynamics and statistical mechanics, as well as anyone seeking to deepen their understanding of Liouville's theorem and its applications in irreversible processes.

annetjelie
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I'm having difficulties understanding what liouville's theorem is all about. I was "meditating" over an adiabatic free expansion and i got stuck because of a contradiction in my reasoning so it seems i still don't have a clue what liouville actually wants me to understand :)

So imagine an ideal gas in volume V that adiabatically expands (free) to a volume 2V. This means more microstates.

Then my contradiction: liouville says that the volume of the microstates (of the macrostate) in phase space under hamiltonian flow stays the same. And as the energy of an ideal gas after free expansion also remains the same, this would mean that the volume of the microstates corresponding to the first macrostate would stay the same
So i get the same "amount" of microstates for the final macrostate which is in contradiction with the above statement and is just plain wrong

Clearly there's a fundamental flaw in my reasoning, and it would mean a lot if someone could give me a nudge in the right direction! Thanks!

*edit* okay i meditated a little more and i guess it's because you have to find a reversible path between the initial and final state because it's an irreversible process, so you can't use Q=0 (and then E won't be constant). Is this right?
 
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So imagine an ideal gas in volume V that adiabatically expands (free) to a volume 2V. This means more microstates.

If the number of gas atoms/molecules N is contant, why would the number of microstates increase?

What is the definition of microstate.
 
Doesn't the number of macrostates increase because V increases?
 
arnesmeets said:
Doesn't the number of macrostates increase because V increases?
that's what i thought. If you expand a gas to a volume twice its initial volume than the phase space of that gas will enlarge by a factor 2^N with N the number of particles, and so there are more possible microstates. No??
 


Your doubt is correct. Liouville theorem does not hold for irreversible process, like adiabatic expansion. Because the phase space is not conserved, as you know, the entropy change is positive.
 

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