Undergrad What Determines the Maximum Number of Microstates at Equilibrium?

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SUMMARY

The maximum number of microstates at equilibrium for two non-interacting systems is determined by the product of their individual microstates, expressed as ##\Omega(E_1, E_2) = \Omega(E_1) \Omega(E_2)##. This relationship arises because each microstate of the first system can be paired with each microstate of the second system, analogous to rolling two six-sided dice where the total outcomes equal the product of their individual outcomes. Therefore, at equilibrium, the total number of accessible microstates is maximized by this multiplicative relationship.

PREREQUISITES
  • Understanding of statistical mechanics concepts
  • Familiarity with microstates and macrostates
  • Knowledge of non-interacting systems in thermodynamics
  • Basic probability theory
NEXT STEPS
  • Research the implications of microstates in statistical mechanics
  • Study the concept of entropy and its relation to microstates
  • Explore the principles of thermodynamic equilibrium
  • Learn about the Boltzmann distribution and its applications
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This discussion is beneficial for students and professionals in physics, particularly those studying thermodynamics and statistical mechanics, as well as researchers interested in the behavior of non-interacting systems at equilibrium.

I_laff
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## \Omega(E_1)## is the number of microstates accessible to a system when it has an energy ##E_1## and ##\Omega(E_2)## is the number of microstates accessible to the system when it has an energy ##E_2##. I understand that each microstate has equal probability of being occupied, but could someone explain at equilibrium why ##\Omega(E_1)\Omega(E_2)## is the maximum number of microstates?
 
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Such a product term means the total number of states

##\Omega(E_1,E_2)=\Omega(E_1)\Omega(E_2)##

of two non-interacting systems with energies ##E_1## and ##E_2##, respectively.
 
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Lord Jestocost said:
Such a product term means the total number of states

##\Omega(E_1,E_2)=\Omega(E_1)\Omega(E_2)##

of two non-interacting systems with energies ##E_1## and ##E_2##, respectively.
Apologies if the answer is obvious, but why does ## \Omega(E_1)\Omega(E_2) ## give the total number of states?
 
The first system may be in any of ##\Omega(E_1)## states; for any of these the second may be in any of ##\Omega(E_2)## states. It’s really no different than a system of consisting of two standard six-sided dice: the first may be in any of six states, the second may be in any of six states, so there are six times six equals thirty-six possible states for the two-dice system.
 
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Nugatory said:
The first system may be in any of ##\Omega(E_1)## states; for any of these the second may be in any of ##\Omega(E_2)## states. It’s really no different than a system of consisting of two standard six-sided dice: the first may be in any of six states, the second may be in any of six states, so there are six times six equals thirty-six possible states for the two-dice system.
Ah of course. Thanks, that cleared things up.
 

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